Question 15 Marks
Find the equation of the hyperbola whose
Vertices are (-8, -1) and (-4, 4) and focus is (17, -1)
AnswerThe center of the hyperbola is the m id-point of the line joining the two foci. So, the coordinates of the centre are $\Big(\frac{16-8}{2},\frac{-1-1}{2}\Big)$ i.e, (4, 1). Let $\text{2a}$ and $\text{2b}$ be the lenght of transverse and conjugate axes and let e be the eccentricity. Then, the equation of the hyperbola is $\frac{\text{(x}-4)^{2}}{\text{a}^{2}}-\frac{(\text{y}+1)^{2}}{\text{b}^{2}}=1$ Now, distance between two vertices = $2\text{a}$ $\therefore$$\sqrt{(16+8)^{2}+(-1+1)^{2}}=2\text{ae}$ [$\because$ vertices = (-8, 1) and (16, -1) $\Rightarrow24=\text{2a}$ $\Rightarrow\text{a}=12$ $\Rightarrow\text{a}^{2}=144$ and, the distance between the focus and vertex is =$\text{ae}-\text{a}$ $\therefore$ $\sqrt{(17-16)^{2}+(-1+1)^{2}}=\text{ae}$ [$\because$ Focus = (17, -1) and vertex = (16,-1)] $\Rightarrow\sqrt{1}^{2}=\text{ae}-\text{a}$ $\Rightarrow\text{ae}-\text{a}=1$ $\Rightarrow12\times\text{e}-12 = 1$$\Big[$$\because\text{a}=12$$\Big]$ $\Rightarrow12\text{e}-1+12$ $\Rightarrow\text{e}=\frac{13}{12}$ $\Rightarrow\text{e}^{2}=\frac{169}{144}$ Now, $\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$ $=(12)^{2}\Big(\frac{169}{144}-1\Big)$ $\Big[\because\text{a} = 12$ and$\text{e}=\frac{13}{12}\Big]$ $=144\times\Big(\frac{169-144}{144}\Big)$ $=144\times\frac{25}{144}$ $=25$ Putting $\text{a}^{2}=144$ and $\text{b}^{2}=25$ in equation {i}, we get $\frac{(\text{x}-4)^{2}}{144}-\frac{(\text{y}+1)^{2}}{25}=1$ $\Rightarrow\frac{25(\text{x}-4)^{2}-144(\text{y}+1)^{2}}{3600}=1$ $\Rightarrow25\text{[x}^{2}+16-8\text{x}]-144[\text{y}^{2}+1+2\text{y}]=3600$ $\Rightarrow25\text{x}^{2}+400-200\text{x}-144\text{y}^{2}-144-288\text{y}=3600$ $\Rightarrow25\text{x}^{2}-144\text{y}^{2}-200\text{x}-288\text{y}+256=3600$ $\Rightarrow25\text{x}^{2}-144\text{y}^{2}-200\text{x}-288\text{y}-3344=0$ This is the equation of the required hyperbola.
View full question & answer→Question 25 Marks
If P is any point on the hyperbola whose axis are equal, prove that SP . SP =$\text{CP}^{2}$
AnswerFor a hyperbola if the lenght of semi transverse and semi conjugate axes are equal.
Then $\alpha\text{=b}$
Equation of the given hyperbola is
$\text{x}^{2}-\text{y}^{2}=\alpha^{2}.....(1)$
Then e $=\sqrt{2}, \text{C}=(0, 0), S=(\sqrt{2\text{a}}, 0), S=(-\sqrt(\text{2a}, 0)$
Let coordinates of any point P on hyperbola be $ (\alpha, \beta). $ Since P lies on (1)
? $\alpha-\beta^{2}=\alpha^{2}......(2)$
Now $\text{SP}^{2}.\text{SP}^{2}$ $=(2\alpha^{2}+\alpha^{2}+\beta^{2})^{2}-8\text{a}^{2}\alpha^{2}$
$=4\alpha^{4}+4\alpha^{2}(\alpha^{2}+\beta^{2})+(\alpha^{2}+\beta^{2})-8\text{a}^{2}\text{a}^{2}$
$=4\text{a}^{2}(\alpha^{2}-2\alpha^{2})+4\text{a}^{2}(\alpha^{2}+\beta^{2})+(\alpha^{2}+\beta^{2})^{2}$
$=4\alpha^{2}(\alpha^{2}-\beta^{2}-2\alpha^{2})+4\text{a}^{2}(\alpha^{2}+\beta^{2})+(\alpha^{2+}\beta^{2})^{2}$
$=(\alpha^{2}+\beta^{2})^{2}=\text{CP}^{4}$
$\text{SP. }\text{SP}=\text{CP}^{2}$
View full question & answer→Question 35 Marks
Find the equation of the hyperbola whoseFoci are (6, 4) and (-4, 4) and eccentricity is 2.
AnswerThe center of the hyperbola is the m id-point of the line joining the two foci.
So, the coordinates of the centre are $\Big(\frac{6-4}{2},\frac{4+4}{2}\Big)$ i.e, (1, 4).
Let $\text{2a}$ and $\text{2b}$ be the lenght of transverse and conjugate axes and let e be the eccentricity.
Then, the equation of the hyperbola is
$\frac{\text{(x}-1)^{2}}{\text{a}^{2}}-\frac{(\text{y}-4)^{2}}{\text{b}^{2}}=1$ ---(i)
Now, distance between two foci = $2\text{ae}$
$\Rightarrow\sqrt{(6+4)^{2}+(4-4)^{2}}=2\text{ae}$ [$\because$ foci = (6, 4) and (-4, 4)]
$\Rightarrow\sqrt{(10)^{2}}=2\text{ae}$
$\Rightarrow10=2\text{ae}$
$\Rightarrow\text{2ae}=10$
$\Rightarrow\text{2a}\times2=10$ [$\because\text{e}=2$]
$\Rightarrow\text{a}=\frac{10}{4}$
$\Rightarrow\text{a}=\frac{5}{2}$
$\Rightarrow\text{a}^{2}=\frac{25}{4}$
Now,
$\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$
$\Rightarrow\text{b}^{2}=\frac{25}{4}(2^{2}-1)$
$=\frac{25}{4}(4-1)$
$=\frac{25}{4}\times3=\frac{75}{4}$
Putting $\text{a}^{2}=\frac{25}{4}$ and $\text{b}^{2}$ = $\frac{75}{4}$ in equation (i), we get
$\frac{\text{(x}-1)^{2}}{\frac{25}{4}}-\frac{(\text{y-4)}^{2}}{\frac{75}{4}}=1$
$\Rightarrow\frac{4(\text{x-1)}^{2}}{25}-\frac{4(\text{y-4)}^{2}}{75}=1$
$\Rightarrow\frac{4\times3(\text{x-1)}^{2}-4\text{(y}-4)^{2}}{75}=1$
$\Rightarrow12(\text{x}-1)^{2}-4(\text{y}-4)^{2}=75$
$\Rightarrow12\big[\text{x}^{2}+1-2\text{x}\big]-4\big[\text{y}^{2}+16-8\text{y}\big]=75$
$\Rightarrow12\text{x}^{2}+12-24\text{x}-4\text{y}^{2}-64+32\text{y}=75$
$\Rightarrow12\text{x}^{2}-4\text{y}^{2}-24\text{x}+32\text{y}-52-75=0$
$\Rightarrow12\text{x}^{2}-4\text{y}^{2}-24\text{x}+32\text{y}-127=0$
This is the equation of the required hyperbola.
View full question & answer→Question 45 Marks
Find the center, eccentricity, foci and directrices of the hyperbola
$\text{x}^{2}-3\text{y}^{2}-2\text{x}=8.$
AnswerWe have,
$\text{x}^{2}-3\text{y}^{2}-2\text{x}=8$
$\Rightarrow\text{x}^{2}-2\text{x}-3\text{y}^{2}=8$
$\Rightarrow\text{x}^{2}-2\text{x}+1-1-3\text{y}^{2}=8$
$\Rightarrow(\text{x}-1)^{2}-1-3\text{y}^{2}=8$
$\Rightarrow\frac{(\text{x}-1)^{2}}{9}-\frac{3\text{y}^{2}}{9}=1$
$\Rightarrow\frac{(\text{x}-1)^{2}}{9}-\frac{\text{y}^{2}}{3}=1$ ----(i)
Shifting the origin at $(1, 0)$ without rotating the axes and denoting the new coordinates w. r. t these axes by X and Y, we have,
$\text{x}=\text{x}+1$ and $\text{y}=\text{y}$ ----(ii)
Using these relations, equation (i) reduces to
$\frac{\text{x}^{2}}{9}-\frac{\text{y}^{2}}{3}=1$ ---(iii)
This is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1$, where $\text{a}^{2}=9$ and $\text{b}^{2}=3.$ so,
We have,
center: The coordinates of the center w.r.t the new axes are $(\text{x}=0, \text{y}=0)$
Putting X $=0$ and $\text{y}=0$ in equation (ii), we get
$\text{x}=1$ and $\text{y}=0.$
So, the coordinates of the centre w.r.t the old axes $(1, 0).$
Eccentricity: The eccentricity e is given by
$\text{e}=\sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$
$=\sqrt{1+\frac{3}{9}}$
$=\sqrt{1+\frac{1}{3}}$
$=\sqrt{\frac{4}{3}}$
$=\frac{2}{\sqrt{3}}$
$=\frac{2\times\sqrt{3}}{\sqrt{3\times\sqrt{3}}}$
$=\frac{2\sqrt{3}}{3}$
Foci: The coordinates of the w.r.t the new axes are $(\text{x}-\pm\text{ae}, \text{y}=0)$ i.e.,$(\text{x}-2\sqrt{3},\text{y}=0$
Putting X $-\pm2\sqrt{3}$ and $\text{y}-0$ in equation (ii), we get
$\text{x}-\pm2\sqrt{3}+1$ and $\text{y}-0$
$\Rightarrow\text{x}=1\pm2\sqrt{3}$ and $\text{y}-0$
So, the coordinates of foci w.r.t the old axes are $(1\pm2\sqrt{3, 0})$
Directrices: The equations of the directrices w.r.t the new axes are
$\text{x}-\pm\frac{\text{a}}{\text{e}}$ i,e.,$\text{x-}\pm\frac{\frac{3}{2\sqrt{3}}}{3}-\pm\frac{9}{2\sqrt{3}}$
Putting X $=\pm\frac{9}{2\sqrt{3}}$ in equation (ii), we get
$\text{x}-\pm\frac{9}{2\sqrt{3}}+1$
$\Rightarrow\text{x}=\pm\frac{9}{2\sqrt{3}}$
View full question & answer→Question 55 Marks
Find the eccentricity of the hyperbola, the length of whose conjugate axis is $\frac{3}{4}$ of the length of transvers axis.
AnswerLet $\text{2a}$ and $\text{2b}$ be the transverse and conjugate axes and e be the eccentricity, Then,
The leng of conjugate axis $=\frac{3}{4}$ [leng the of transverse axis]
$\Rightarrow2\text{b}=\frac{3}{4}\times(2\text{a})$
$\Rightarrow\frac{\text{b}}{\text{a}}=\frac{3}{4}$
$\Rightarrow\frac{\text{b}^{2}}{\text{a}^{2}}=\frac{9}{16}$
Now,
$\text{e}=\sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$
$=\sqrt{1+\frac{9}{16}}$
$=\sqrt{\frac{25}{16}}$
$=\frac{5}{4}$
Hence, e $=\frac{5}{4}$
View full question & answer→Question 65 Marks
Find the equation of the hyperbola whose Vertices are at $(0 \pm 7)$ and foci at $\big(0, + \frac{28}{3}\big).$
AnswerSince, the vertices are on y-axis, so let the equation of the requried hyperbola is
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}} = -1 ---(\text{i})$
The coordinates of its vertices and foci are $(0, \pm \text{b})$ and $(0, \pm \text{b})$ respectively.
$\therefore\text{b} = 7$ $\big[\because$ vertices = $(0, \pm 7)\big]$
$\Rightarrow\text{b}^{2}-49$
and,
$\text{be} = \frac{28}{3}$ $\Big[\because$ Foci = $\big(0, \pm\frac{28}{3}\big)\Big]$
$\Rightarrow7 \times \text{e} = \frac{28}{3}$
$\Rightarrow\text{e} - \frac{4}{3}$
$\Rightarrow\text{e}^{2}-\frac{16}{9}$
Now,
$\text{a}^{2}-\text{b}^{2}\big(\text{e}^{2}-1\big)$
$\Rightarrow\text{a}^{2} = 49\big(\frac{16}{9}-1\big)$
$\Rightarrow\text{a}^{2} = 49 \times\frac{7}{9}$
$\Rightarrow\text{a}^{2} = \frac{343}{9}$
Putting $\text{a}^{2} - \frac{343}{9}$ and $\text{b}^{2} - 49$ in equation (i), we get
$\frac{343\text{ x}^{2}}{9}-\frac{\text{y}^{2}}{49} = -1$
This is the equation of the required hyperbola.
View full question & answer→Question 75 Marks
In each the following find the equation of the hyperbola satisfying the given conditions:Foci $(\pm3\sqrt{5}, 0)$, the latus-rectum = 8 [NCERT]
AnswerSince, the vertices line on x-axis, so let the equation of the required hyperbola be
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1---(\text{i})$
The lenght of conjugater axis of the required hyperbola is 8.
$\therefore\frac{2\text{b}^{2}}{\text{a}}=8$
$\Rightarrow\text{b}^{2}=\frac{8}{2}\times\text{a}$
$\Rightarrow\text{b}^{2}=4\text{a}---(\text{ii})$
Now,
This coordinates of foci of the required hyperbola is $(\pm\text{ae}, 0)$
$\therefore\text{ae}=3\sqrt{5}$ $\big[\because$ Foci = $\big(\pm3\sqrt{5},0\big)\big]$
$\Rightarrow\text{e}=\frac{3\sqrt{5}}{\text{a}}$
$\Rightarrow\text{e}^{2}=\frac{45}{\text{a}^{2}}---(\text{iii})$
Now,
$\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$
$\Rightarrow4\text{a}=\text{a}^{2}\text{e}^{2}-\text{a}^{2}$
$\Rightarrow4\text{a}=\text{a}^{2}\times\frac{45}{\text{a}^{2}}-\text{a}^{2}$
$\Rightarrow4\text{a}=45-\text{a}^{2}$
$\Rightarrow\text{a}^{2}+4\text{a}-45=0$
$\Rightarrow\text{a}^{2}+9\text{a}-5\text{a}-45=0$
$\Rightarrow\text{a}(\text{a}+9)-5(\text{a}+9)=0$
$\Rightarrow(\text{a}-5)(\text{a}+9)=0$
$\Rightarrow\text{a}=5$ $[\because\text{a+9}\not=0]$ $$
$\Rightarrow\text{a}^{2}=25$
$\Rightarrow\text{b}^{2}=4\times5$ [Using equation (ii)]
$\Rightarrow\text{b}^{2}=20$
Putting $\text{a}^{2}=25$ and $\text{b}^{2}=20$ in equation (i), we get
$\frac{\text{x}^{2}}{25}-\frac{\text{y}^{2}}{20}=1$
Hence, the equation of the required hyperbola is
$\frac{\text{x}^{2}}{25}-\frac{\text{y}^{2}}{20}=1$.
View full question & answer→Question 85 Marks
In each of the followin find the equation of the hyperbola satisying the given conditions: vertices $(0, \pm3)$, foci $(0, \pm5)$ [NCERT]
AnswerSince, the vertices line on y-axis, So let the equation of the required hyperbola be $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=-1 ---(\text{i})$
The coordinates of its vertices and foci are $(0, \pm\text{b})$ and $(0, \pm\text{be})$ respectively.
$\therefore\text{b}=3$ $[\because$ vertices = $(0, \pm)]$
$\Rightarrow\text{b}^{2}-9$
and, $\text{be} = 5$ $[\because$ Foci = $(0, \pm5)]$
$\Rightarrow\text{e}\times3=5$
$\Rightarrow\text{e}\times3=5$
$\Rightarrow\text{e}=\frac{5}{3}$
$\Rightarrow\text{e}^{2}=\frac{25}{9}$
Now,
$\text{a}^{2}-\text{b}^{2}(\text{e}^{2}-1)$
$\Rightarrow\text{a}^{2}-9(\frac{25}{9}-1)$
$= 9\times(\frac{25-9}{9})$
$=9\times\frac{16}{9}$
$=16$
Putting $\text{a}^{2}-16$ and $\text{b}^{2}-9$ in equation (i), we get
$\frac{\text{x}^{2}}{16}-\frac{\text{y}^{2}}{9}=-1$
Hence, the equation of the required hyperbola is
$\frac{\text{x}^{2}}{16}-\frac{\text{y}^{2}}{9}=-1$
View full question & answer→Question 95 Marks
Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
the distance between the foci = 16 and eccentricity $=\sqrt{2}$
AnswerLet the equation of the hyperbola be
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1$ ----(i)
$\Rightarrow\text{2ae}=16$ $[\because$ Distance between foci = $2\text{ae}]$
$\Rightarrow\text{ae}=8$
$\Rightarrow\text{a}\times\sqrt{2}=8$ [$\because\text{e}=\sqrt{2}$]
$\Rightarrow\text{a}=\frac{8}{\sqrt{2}}$
$\Rightarrow\text{a}^{2}=\frac{64}{2}=32$
Now,
$\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$
$=32\Big((\sqrt{2)}^{2}-1\Big)$
$=32\times(2-1)$
$=32$
Putting $\text{a}^{2}$ =32 and $\text{b}^{2}$ = 32 in equation (i), we get
$\frac{\text{x}^{2}}{32}-\frac{\text{y}^{2}}{32}=1$
$\Rightarrow\text{x}^{2}-\text{y}^{2}=32$
Hence, the equation of the required huperbola is $\text{x}^{2}-\text{y}^{2}=32.$
View full question & answer→Question 105 Marks
If the distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$, then obtain its equation.
AnswerEccentricity = $\text{e}=\sqrt{2}$ Distance between foci is $2\text{ae}=16$ $2\alpha\sqrt{2}=16$ $\alpha=\frac{16}{2\sqrt{2}}=4\sqrt{2}$ $\text{e}=\frac{\sqrt{\alpha^{2}+\beta^{2}}}{\alpha}$ $\sqrt{2}=\frac{\sqrt{32+\beta^{2}}}{4\sqrt{2}}$ $8=\sqrt{32+\beta^{2}}$ $64=32+\beta^{2}$ $\beta(2)=32$ Equation of hyperbola is $\frac{\text{x}^{2}}{32}-\frac{\text{y}^{2}}{32}=1$ Rewriting we get, $\text{x}^{2}-\text{y}^{2}=32$
View full question & answer→Question 115 Marks
In each of the following find the equations of the hyperbola satisfying the given conditions:Vertices $(\pm2, 0)$, foci $(\pm3, 0)$ [NCERT]
AnswerLet the equation of hyperbola be
$\frac{\text{x}^{2}}{\text{a}^{2}}-1---(\text{i})$
The coordinates of its vertices and foci are $(\pm\text{a, 0})$ and $(\pm\text{ae}, 0)$ respectively.
$\therefore\text{a}= 2$ [$\because$ vertices = $(\pm2, 0)$]
$\Rightarrow\text{a}^{2}-4$
and,
$\text{ae}=3$ [$\because$ Foci = $(\pm3, 0)$]
$\Rightarrow2\times\text{e}=3$ [$\because\text{a}=2$]
$\Rightarrow\text{e}=\frac{3}{2}$
Now,
$\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$
$\Rightarrow\text{b}^{2}-2^{2}\Big[(\frac{3}{2})^{2}-1\Big]$
$\Rightarrow\text{b}^{2}-4\Big[\frac{9}{4}-1\Big]$
$\Rightarrow\text{b}^{2}=4\Big[\frac{9-4}{4}\Big]$
$=4\times\frac{5}{4}$
$=5$
Putting $\text{a}^{2}-4$ and $\text{b}^{2}-5$ in equation (1), we get
$\frac{\text{x}^{2}}{4}-\frac{\text{y}^{2}}{5}=1$
Hence, the equation of the required hyperbola is $\frac{\text{x}^{2}}{4}-\frac{\text{y}^{2}}{5}-1$.
View full question & answer→Question 125 Marks
Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
conjugate axis is 5 and the distance between foci = 13
AnswerLet the aquation of the hyperbola be
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1\ \dots(1)$
Then,
The lenght of the conjugate axis $=2\text{b}$
$\therefore2\text{b}=5$ [$\because$ Conjugate axis = 5$\big]$
$\Rightarrow\text{b}=\frac{5}{2}$
$\Rightarrow\text{b}^{2}=\frac{25}{4}$
And, the distance between foci $=2\text{ae}$
$\therefore2\text{ae}=13$ [$\because$ The distance between foci is 13$\big]$
$\Rightarrow\text{a}^{2}\text{e}^{2}=\frac{169}{4}$
Now,
$\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$
$\Rightarrow\frac{25}{4}=\text{a}^{2}\text{e}^{2}-\text{a}^{2}$
$\Rightarrow\frac{25}{4}=\frac{169}{4}-\text{a}^{2}$
$\Rightarrow\text{a}^{2}=\frac{169}{4}-\frac{25}{4}$
$\Rightarrow\text{a}^{2}=\frac{169-25}{4}$
$\Rightarrow\text{a}^{2}=\frac{144}{4}=36$
Putting $\text{a}^{2} = 36$ and $\text{b}^{2}=\frac{25}{4}$ in equation (i), we get
$\frac{\text{x}^{2}}{36}-\frac{\frac{\text{y}^{2}}{25}}{4}=1$
$\Rightarrow\frac{\text{x}^{2}}{36}-\frac{4\text{y}^{2}}{25}=1$
$\Rightarrow25\text{x}^{2}-144\text{y}^{2}=900$
Hence, the equation of the required hyperbola is $25\text{x}^{2}-144\text{y}^{2}=900.$
View full question & answer→Question 135 Marks
In each of the following find the equation of the hyperbola satisfying the given conditions
foci $(0, \pm12), $ latus-rectum=36 [NCERT]
AnswerSince, the vertices line on x-axies, so let the equation of the required hyperbola be $\frac{\text{x}^{2}}{\alpha^{2}}-\frac{\text{y}^{2}}{\beta^{2}}=1---(\text{i})$ The length of the latus-rectum of the required hyperbola is 36. $\frac{2\alpha^{2}}{\beta}=36$ $\alpha^{2}=186---(\text{ii})$ Now, The coordinates of the required hyperbola is ($0, \pm\text{be}$). $\beta\text{e}=12 $ $\text{e}=\frac{12}{\beta}$ $\text{e}^{2}=\frac{144}{\beta^{2}}$ Now, $\alpha^{2}=\beta^{2}(\text{e}^{2}-1)$ $186=\beta^{2}\big(\frac{144}{\beta^{2}}-1\big)$ $186=144-\beta^{2}$ $\beta^{2}+186-144=0$ $(\beta-6)(\beta+24)=0$ $\beta_2=6, -24$ Consider the positive value of $\beta=6$ On putting $\beta^{2}=36, \alpha^{2}=18(6)=108$in equation (i), we get $\frac{\text{x}^{2}}{108}-\frac{\text{y}^{2}}{36}=-1$ $\frac{\text{x}^{2}-3\text{y}^{2}}{108}=-1$ $\text{x}^{2}-3\text{y}^{2}=-108$ $3\text{y}^{2}-\text{x}^{2}=108$ Therefore, the equation of the hyperbola is $3\text{y}^{2}-\text{x}^{2}=108$
View full question & answer→Question 145 Marks
Find the equation of the hyperbola whoseFoci at $(\pm2, 0)$ and eccentricity is $\frac{3}{2}$. [NCRT EXEMPLAR]
AnswerThe foci of the hyperbola are $(\pm2, 0).$
$\therefore$
$\text{ae} = 2$
$\Rightarrow\text{a}=2\times\frac{2}{3}=\frac{4}{3}$
$\Rightarrow\text{a}^{2}=\frac{16}{9}$
Now,
$\text{(ae)}^{2}=\text{a}^{2}+\text{b}^{2}$
$\Rightarrow(2)^{2}=\big(\frac{4}{3}\big)^{2}+\text{b}^{2}$
$\Rightarrow4-\frac{16}{9}=\text{b}^{2}$
$=\text{b}^{2}=\frac{20}{9}$
Therefore, the equation of the hyperbola is given by
$\frac{9\text{x}^{2}}{16}-\frac{9\text{y}^{2}}{20}=1$
$\Rightarrow\frac{\text{x}^{2}}{4}-\frac{\text{y}^{2}}{5}=\frac{4}{9}$
View full question & answer→Question 155 Marks
In each the following find the equation of the hyperbola satisfying the given conditions:
Foci $(\pm4, 0), $ latus-rectum = 12 [NCERT]
AnswerSince, the vertices line on x-axis, so let the equation of the required hyperbola be
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1---(\text{i})$
The lenght of the latus-rectum of the required hyperbola is 12.
$\therefore\frac{2\text{b}^{2}}{\text{a}}=12$
$\Rightarrow\text{b}^{2}=6\text{a}---(\text{ii})$
Now,
The coordinates of foci of the required hyperbola is $(\pm\text{ae, 0})$
$\therefore\text{ae}=4$
$\Rightarrow\text{e}=\frac{4}{\text{a}}$
$\Rightarrow\text{e}^{2}=\frac{16}{\text{a}^{2}}---(\text{iii})$
Now,
$\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$
$\Rightarrow6\text{a}=\text{a}^{2}\text{e}^{2}-\text{a}^{2}$
$\Rightarrow6\text{a}=\text{a}^{2}\times\frac{16}{\text{a}^{2}}-\text{a}^{2}$
$\Rightarrow6\text{a}=16-\text{a}^{2}$
$\Rightarrow\text{a}^{2}+6\text{a}-16=0$
$\Rightarrow\text{a}^{2}+8\text{a}-2\text{a}-16=0$
$\Rightarrow\text{a}(\text{a}+8)-2(\text{a}+8)=0$
$\Rightarrow(\text{a}+8)(\text{a}-2)$
$\Rightarrow(\text{a}-2)=0$ $\begin{bmatrix}\because\text{ lenght cannot be negative}\\\therefore\ \text{a+8} \neq0 \end{bmatrix}$
$\Rightarrow\text{a}=2$
$\Rightarrow\text{a}^{2}=4$
$\Rightarrow\text{b}^{2}=6\times2=12$[Using equation (ii)]
Putting $\text{a}^{2}=4$ and $\text{b}^{2}=12$ in equation (i), we get
$\frac{\text{x}^{2}}{4}-\frac{\text{y}^{2}}{12}=1$
Hence, the equation of the required hyperbola is
$\frac{\text{x}^{2}}{4}-\frac{\text{y}^{2}}{12}=1.$
View full question & answer→Question 165 Marks
Find the equation of the hyperbola whose
Focus is at $(4, 2)$ centre at $(6, 2)$ and $e = 2.$
AnswerThe equation of the hyperbola with centre $(X_0, Y_0)$ is given by
$\frac{\text{(x}-\text{x}_0)}{\text{a}^{2}}-\frac{(\text{y}-\text{y}_0)^{2}}{\text{b}^{2}}=1$
Focus = $(\text{ae}+\text{x}_0, \text{y}_0)$
$\therefore\text{ae} = -2$
$\Rightarrow\text{a} = -1$
$\text{b}^{2} = (-2)^{2}-\text{a}^{2}$
$\Rightarrow\text{b}^{2}=(-2)^{2}-(-1)^{2}$
$\Rightarrow\text{b}^{2}=3$
$\Rightarrow\frac{\text{(x}-6)^{2}}{1}-\frac{(\text{y}-2)^{2}}{3}=1$
$\Rightarrow3\text{(x}-6)-(\text{y}-2)^{2}=3$
View full question & answer→Question 175 Marks
Find the equation of the hyperbola whose
vertices are at $(\pm6, 0)$ and one of the directrices is x = 4
AnswerThe vertices of the hyperbola are $(\pm6, 0)$ $\therefore\text{a}=6$ $\Rightarrow\text{a}^{2}=36$ Now, $\text{x}=4$ $\frac{\text{a}}{\text{e}}=4$ $\Rightarrow\text{e}=\frac{3}{2}$ $\big[\because\text{a}=6\big]$ Now, $\text{(ae)}^{2}=\text{a}^{2}+\text{b}^{2}$ $\Rightarrow\big(6\times\frac{3}{2}\big)^{2}=6^{2}+\text{b}^{2}$ $\Rightarrow81-36=\text{b}^{2}$ $\Rightarrow\text{b}^{2}=45$ Therefore, the equqtion of the hyperbola is $\frac{\text{x}^{2}}{36}-\frac{\text{y}^{2}}{45}=1.$
View full question & answer→Question 185 Marks
In each the following find the equation of the hyperbola satisfying the given conditions:
Foci $(0, \pm13), $ conjugate axis = 24
AnswerSince, the vertices line on x-axis, so let the equation of the required hyperbola be
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=-1---(\text{i})$
The lenght of conjugater axis of the required hyperbola is 24.
$\therefore2\text{a}=24$ [$\because$ conjugate axis is $2\text{a}$]
$\Rightarrow\text{a}=\frac{24}{2}=12$
$\Rightarrow\text{a}^{2}-144$
This coordinates of foci of the required hyperbola is $(0,\pm\text{be})$
$\therefore2\text{a}=24$ [$\because$ conjugate axis is $2\text{a}$]
$\Rightarrow\text{a}=\frac{24}{2}=12$
$\Rightarrow\text{a}^{2}-144$
This coordinates of foci of the required hyperbola is $(0, \pm\text{be})$
$\therefore\text{be}=13$
$\text{b}^{2}\text{e}^{2}=169$
Now,
$\text{a}^{2}=\text{b}^{2}(\text{e}^{2}-1)$
$\Rightarrow144=\text{b}^{2}\text{e}^{2}-\text{b}^{2}$
$\Rightarrow144=169-\text{b}^{2}$
$\Rightarrow\text{b}^{2}-169-144-25$
Putting $\text{a}^{2}=144$ and $\text{b}^{2}=25$ in equation (i), we get
$\frac{\text{x}^{2}}{144}-\frac{\text{y}^{2}}{25}=1-1$
Hence, the equation of the required hyperbola is
$\frac{\text{x}^{2}}{144}-\frac{\text{y}^{2}}{25}=-1.$
View full question & answer→Question 195 Marks
Find the equation of the hyperbola whose,Focus is (2, 2) directrix is $\text{x+y}=\text{9}$ and eccentricity = 2
AnswerLet S (2, 2) be the focus and P (x, y) be a point on the hyperbola, Draw PM perpendicular from P on the directrix, Then, By definition$\text{sP}=\text{ePM}$
$\Rightarrow\text{sP}^{2}=\text{e}^{2}\text{PM}^{2}$ $\Rightarrow(\text{x}-2)^{2}+(\text{y-2})^{2}=2^{2}\Bigg[\frac{\text{x}+\text{y}-9}{\sqrt{1^{2}+1^{2}}}\Bigg]^{2}$ $\Rightarrow\text{x}^{2}+4-4\text{x}+\text{y}^{2}+4-4\text{y}=\frac{4[\text{x+y}-9]^{2}}{2}$ $\Rightarrow\text{x}^{2}+\text{y}-4\text{x}-\text{4y}+8=2[\text{x+y}-9]^{2}$ $\Rightarrow\text{x}^{2}+\text{y}^{2}-4\text{x}-4\text{y}+8\\=2\Big[\text{x}^{2}+\text{y}^{2}+(-9)^{2}+2\times\text{x}\times\text{y+2}\times\text{y}\times(-9)+2\times(-9)\times\text{x}\Big]$ $\Rightarrow\text{x}^{2}+\text{y}^{2}-4\text{x}-4\text{y}+8=2\Big[\text{x}^{2}+\text{y}^{2}+81+2\text{xy}-18\text{y}+18\text{x}\Big]$ $\Rightarrow\text{x}^{2}+\text{y}^{2}-4\text{x}-4\text{y}+8=\big[2\text{x}^{2}+2\text{y}^{2}+162+4\text{xy}-36\text{y}-36\text{x}$ $\Rightarrow2\text{x}^{2}-\text{x}^{2}+2\text{y}^{2}-\text{y}^{2}+4\text{xy}-36\text{x}+4\text{x}-36\text{y}+4\text{y}+162-8=0$ $\Rightarrow\text{x}^{2}+\text{y}^{2}+4\text{xy}-32\text{x}-32\text{y}+154=0$ This is the required equation of the hyperbola.
View full question & answer→Question 205 Marks
Find the eccentricity, coordinates of the foci, equation of the directrices and lenght of the latus-rectum of the hyperbola
$9\text{x}^{2}-16\text{y}^{2}=144$
AnswerWe have,
$9\text{x}^{2}-16\text{y}^{2}=144$
$\Rightarrow \frac{9\text{x}^{2}}{144}-\frac{16\text{y}^{2}}{144}=1$
$\Rightarrow\frac{\text{x}^{2}}{16}-\frac{\text{y}^{2}}{9}=1$
This is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1, $ Where $\text{a}^{2}=16$ and $\text{b}^{2}=9$
Eccentricity: The eccentricity e is given by
$\text{e} = \sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$
$= \sqrt{1+\frac{9}{16}}$
$= \sqrt{\frac{25}{16}}$
$= \frac{5}{4}$
Foci: The coordinates of the foci are $(\pm\text{ae, 0})$ i.e, $(\pm5, 0)$
Equations of the directrices; The equations of the directrices are
$\text{x} = \pm\frac{\text{a}}{\text{e}}\text{ i}.\text{e.,}\times = \pm\frac{16}{5}$
$\therefore5\text{x}=\pm16$
$\Rightarrow5\text{x}\mp16 = 0$
Lenght of latus-rectum; The lenght of the latus-rectum
$=\frac{2\text{b}^{2}}{\text{a}}=\frac{2\times9}{4}=\frac{9}{2}$
View full question & answer→Question 215 Marks
Find the equation of the hyperbola whoseFocus is at $(5, 2),$ vertex at $(4, 2)$ and center at $(3, 2)$
AnswerThe equation of the hyperbola with center $(X_{0},Y_0)$ is given by
$\frac{(\text{x}-\text{x}_0)^{2}}{\text{a}^{2}}-\frac{\text{(y}-\text{y}_0)^{2}}{\text{b}^{2}}=1$
Focus $= (\text{ae} + \text{x}_0, \text{y}_0)$
Vertex $=\text{a}+\text{x}_0, \text{y}_0$
$\therefore\text{ae}=2$
and $\text{a}=1$
$\text{b}^{2}(2)^{2}-\text{a}^{2}$
$\Rightarrow\text{b}^{2}=(2)^{2}-(1)^{2}$
$\Rightarrow\text{b}^{2}=3$
$\Rightarrow\frac{\text{(x}-3)^{2}}{1}-\frac{(\text{y}-2)^{2}}{3}=1$
$\Rightarrow3(\text{x}-3)^{2}-(\text{y}-2)^{2}=3$
View full question & answer→Question 225 Marks
Find the equation of the hyperbola whose
focus is (1, 3), directrix is x + y - 1 = 0 and eccentricity = 2
AnswerLet (0, 3) be the focus and p (x, y) be a point a on the hyperbola, Draw PM perpendicular from p on the directrix, then, by definition
$\text{sP}=\text{ePM}$
$\Rightarrow \text{sP}^{2}=\text{e}^{2}\text{PM}^{2}$
$\Rightarrow (\text{x}-0)^{2}+\text{(y}-3)^{2}=2^2\Bigg[\frac{\text{x}+\text{y}-1}{\sqrt{1^{2}+1^{2}}}\Bigg]^{2}$
$\Rightarrow\text{x}^{2}+\text{y}^2+9-6\text{y}=\frac{4[\text{x}+\text{y}-1]^{2}}{2}$
$\Rightarrow\text{x}^{2}+\text{y}^{2}-6\text{y}+9=2\text{(x}+\text{y}-1)^{2}$
$\Rightarrow\text{x}^{2}+\text{y}^{2}-6\text{y}+9=2\bigg[\text{x}^{2}+\text{y}^{2}+(-\text{1})^{2}+2\text{xy}+2\text{xy}\times(-1)+2\times(-1)\times\text{x}\bigg]$
$\Rightarrow\text{x}^{2}+\text{y}^{2}-6\text{y}+9=2\bigg[\text{x}^{2}+\text{y}^{2}+1+2\text{xy}-2\text{y}-2\text{x}\bigg]$
$\Rightarrow\text{x}^{2}+\text{y}^{2}-6\text{y}+9=2\text{x}^{2}+2\text{y}^{2}+2+4\text{xy}-4\text{y}-\text{4x}$
$\Rightarrow2\text{x}^{2}-\text{x}^{2}+2\text{y}^{2}-4\text{xy}-4\text{x}-4\text{y}+6\text{y}+2-9=0$
$\Rightarrow\text{x}^{2}+\text{y}^{2}+4\text{xy}-4\text{x}+2\text{y}-7=0$
This is the requierd equation of the hyperbola.
View full question & answer→Question 235 Marks
Find the equation of the hyperbola whose,
Focus is (1, 1) directrix is $\text{2x}+\text{y}=1$ and eccentricity $= \sqrt{3}$
AnswerLet S (-1, 1) be the focus and P (x, y) be a point on the hyperbola Draw PM perpendicular from P on the directrix, Then, by definition
$\text{sP}=\text{ePM}$
$\Rightarrow\text{SP}^{2}=\text{e}^{2}\text{PM}^{2}$
$\Rightarrow(\text{x+1})^{2}+(\text{y-1})^{2}=(3)^{2}\Bigg[\frac{\text{x}-\text{y}+3}{\sqrt{1^{2}+(-1)^{2}}}\Bigg]^{2}$
$\Rightarrow\text{x}^{2}+1+2\text{x}+\text{y}^{2}+1-2\text{y}=\frac{9[\text{x}-\text{y}+3]^{2}}{2}$
$\Rightarrow2\text{[x}^{2}+\text{y}^{2}+2\text{x}-2\text{y}+2]=9[\text{x}-\text{y}+3]^{2}$
$\Rightarrow2\text{x}^{2}+2\text{y}^{2}+4\text{x}-4\text{y}+4\\=9\Big[\text{x}^{2}(-\text{y)}^{2}+3^{2}+2\times\text{x}\times\text{x}(-\text{y})\times3+2\times3\times\text{x}\Big]$
$\Rightarrow\text{2}\text{x}^{2}+2\text{y}^{2}+4\text{x}-4\text{y}-4=9\Big[\text{x}^{2}(-\text{y})^{2}+9-2\text{xy}-6\text{y}+\text{6x}$
$\Rightarrow2\text{x}^{2}+2\text{y}^{2}+4\text{x}-4\text{y}+4\\=9\text{x}^{2}+9\text{y}^{2}+81-18\text{xy}-54\text{y}+4\text{y}+81-4=0$
$\Rightarrow7\text{x}^{2}+7\text{y}^{2}-18\text{xy}+50\text{x}-50\text{y}+77=0$
This is the required equation of the hyperbola.
View full question & answer→Question 245 Marks
Find the eccentricity, coordinates of the foci, equations of directrices and lenght of the latus-rectum of the hyperbola
$\text{3x}^{2}-\text{y}^{2}=4$
AnswerWe have,
$\text{3x}^{2}-\text{y}^{2}=4$
$\Rightarrow\frac{\text{3x}^{2}}{4}-\frac{\text{y}^{2}}{4}=1$
$\Rightarrow\frac{\frac{\text{x}^{2}}{4}}{3}-\frac{\text{y}^{2}}{4}=1$
$\Rightarrow\frac{\text{x}^{2}}{\Big(\frac{2}{\sqrt{3}}\Big)^{2}}-\frac{\text{y}^{2}}{2^{2}}{}=1$
This is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1$, where a = $\frac{2}{\sqrt{3}}$ and $\text{b}=2$
Eccentricity: The eccentricity e is given by
$\text{e}=\sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$
$=\sqrt{1+\frac{\frac{4}{4}}{3}}$
$=\sqrt{1+3}$
$=\sqrt{4}$
$=2$
Foci: The coordinates of the foci are $(\pm\text{ae, 0})$
$\therefore\pm\text{ae}=\pm\frac{2}{\sqrt{3}}\times2=\pm\frac{4}{\sqrt{3}}$
The coordinates of the foci are $\Big(\pm\frac{4}{\sqrt{3}}, 0\Big)$
Equations of the directirices: The equations of the directrices are
$\text{x}=\pm\frac{\text{a}}{\text{e}}$
$=\pm\frac{\frac{2}{\sqrt{3}}}{2}$
$=\pm\frac{1}{\sqrt{3}}$
$\Rightarrow\sqrt{3\text{x}}\mp1=0$
Latus-rectum: The lenght of the latus-rectum = $\frac{2\text{b}^{2}}{\text{a}},$
$\therefore\frac{\text{2b}^{2}}{\text{a}}=2\times\frac{\frac{4}{2}}{\sqrt{3}}$
$=4\sqrt{3}$
View full question & answer→Question 255 Marks
The equation of the directeix of a hyperbola is x - y + 3 = 0, its focus is ( - 1 , 1 ) and eccentricity 3. find the equation of the hyperbola.
AnswerLet s (-1 , 1) be the focus and p(x, y) be a point on the hyperbola Draw pm perpendicular from p on the directrix, then , by definition.
$\text{sp} = \text{epm}$
$\Rightarrow \text{sp}^{2}= \text{e}^{2}\text{pm}^{2}$
$\Rightarrow \text{( x+1 )}^{2}+( \text{y} - 1 )^{2} = ( 3 )^{2}\Bigg[\frac{\text{x}-\text{y}+3}{\sqrt{1^{2}+( - 1 )^{2}}}\Bigg]$
$\Rightarrow \text{( x )}+1+2\text{x}+\text{y}^{2}+1-2\text{y} = \frac{9[\text{x}-\text{y}+3]^{2}}{2}$
$\Rightarrow2[\text{x}^{2}+\text{y}^{2}+2\text{x}-2\text{y}+2] = 9[\text{x}-\text{y}+3]^{2}$
$\Rightarrow 2\text{x}^{2}+2\text{y}^{2}+4\text{x}-4\text{y}+4 = 9[\text{x}^{2}( -\text{y})^{2}+3^{2}+2\times\text{x}\times(-\text{y})\\\ \ +2\text{x}(-\text{y})+2\times(-\text{y})\times3+2\times3\times\text{x}]$
$\Rightarrow2\text{x}^{2}+2\text{y}^{2}+4\text{x}-4\text{y}-4=9[\text{x}^{2}+\text{y}^{2}+9-2\text{xy}-6\text{y}+6\text{x]}$
$\Rightarrow2\text{x}^{2}+2\text{y}^{2}+4\text{x}-4\text{y}+4=9\text{x}^{2}+9\text{y}^{2}+81\text{xy}\\\ -18\text{xy}-54\text{y}+4\text{y}-54\text{y}+4\text{y}+81-4=0$
$\Rightarrow7\text{x}^{2}+7\text{y}^{2}-18\text{xy}+50\text{x}-50\text{y}+77=0$
This is the required equation of the hyperbola.
View full question & answer→Question 265 Marks
Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:
Conjugate axis is 7 and passes throught the point (3,-2)
AnswerLet the equation of the hyperbola be $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}-1$ ----(i) Then, The lenght of the conjugate axis $-2\text{b}$ $\therefore2\text{b}=7$ [$\because$ Conjugate axis is = 5] $\Rightarrow\text{b}-\frac{7}{2}$ $\Rightarrow\text{b}^{2}-\frac{49}{4}$ ----(ii) The required hyperbola passes throught the point (3, 2). $\therefore$ $\frac{(3)^{2}}{\text{a}^{2}}-\frac{(-2)^{2}}{\text{b}^{2}}=1$ $\Rightarrow\frac{\text{a}}{\text{a}^{2}}-\frac{\frac{\text{y}}{49}}{4}=1$ $\Rightarrow\frac{9}{\text{a}^{2}}-\frac{16}{49}=1$ $\Rightarrow\frac{9}{\text{a}^{2}}-1+\frac{16}{49}$ $\Rightarrow\frac{9}{\text{a}^{2}}=\frac{65}{49}$ $\Rightarrow\text{a}^{2}-\frac{49\times9}{65}$ $\Rightarrow\text{a}^{2}-\frac{441}{65}$ Putting $\text{a}^{2}-\frac{441}{65}$ and $\text{b}^{2}-\frac{49}{4}$ in equation (i), we get $\frac{\frac{\text{x}^{2}}{441}}{65}-\frac{\frac{\text{y}^{2}}{49}}{4}=1$ $\Rightarrow\frac{65\text{x}^{2}}{441}-\frac{4\text{y}^{2}}{49}=1$ $\Rightarrow\frac{65\text{x}^{2}-36\text{y}^{2}}{441}=1$ $\Rightarrow65\text{x}^{2}-36\text{y}^{2}=441$ Hence, the equation of the required hyperbola is $65\text{x}^{2}-36\text{y}-441.$
View full question & answer→Question 275 Marks
Find the eccentricity, coordinates of the foci, equations of directrices and lenght of the latus-rectum of the hyperbola
$2\text{x}^{2}-3\text{y}^{2}=5.$
Answer(v) Equation of the hyperbola:$\text{2x}^{2}-3\text{y}^{2}=5$
This can be rewritten in the following manner:
$\frac{\text{2x}^{2}}{5}-\frac{3\text{y}^{2}}{5}=1$
$\Rightarrow\frac{\text{x}^{2}}{\frac{5}{2}}-\frac{\text{y}^{2}}{\frac{5}{3}}=1$
This is the standard equation of a hyperbola, where $\text{a}^{2}=\frac{5}{2}$ and $\text{b}^{2}=\frac{5}{3}.$
$\Rightarrow\text{b}^{2}=\text{a}^{2}\Big(\text{e}^{2}-1\Big)$
$\Rightarrow\frac{5}{3}=\frac{5}{2}\Big(\text{e}^{2}-1\Big)$
$\Rightarrow\text{e}^{2}-1=\frac{2}{3}$
$\Rightarrow\text{e}^{2}=\frac{5}{3}$
$\Rightarrow\text{e}=\sqrt{\frac{5}{3}}$
Coordinates of the foci are given by $(\pm\text{ae, 0}), $ i.e. $\Big(\pm\frac{5\sqrt{6}}{6}, 0\Big).$
Equation of the directrices:
$\text{x}=\pm\frac{\text{a}}{\text{e}}$
$\text{x}=\pm\frac{\sqrt{\frac{5}{2}}}{\sqrt{\frac{5}{3}}}$
$\Rightarrow\text{x}=\pm\frac{\sqrt{3}}{\sqrt{2}}$
$\Rightarrow\sqrt{\text{2x}}\pm\sqrt{3}=0$
Lenght of the latus rectum of the hyperbola is $\frac{2\text{b}^{2}}{\text{a}}.$
$\Rightarrow\frac{2\times\Big(\frac{5}{3}\Big)}{\sqrt{\frac{5}{2}}}=\frac{10}{3}\sqrt{\frac{2}{5}}$
View full question & answer→Question 285 Marks
Find the equation of the hyperbola whose,
Focus is (2, 1) directrix is $2\text{x}+3\text{y}=1$ and eccentricity = 2
AnswerLet (2, 1) be the focus and P (x, y) be a point on the hyperbola, Draw PM perpendicular from P on the directrix, Then, by definition
$\text{sP}=\text{ePM}$
$\Rightarrow\text{sP}^{2}=\text{e}^{2}\text{PM}^{2}$
$\Rightarrow(\text{x-2)}^{2}+(\text{y}-1)^{2}=2^{2}\Bigg[\frac{2\text{x}+3\text{y}-1}{\sqrt{2^{2}+3^{2}}}\Bigg]^{2}$
$\Rightarrow\text{x}^{2}+4-4\text{x}+\text{y}^{2}+1+2\text{y}=\frac{4[2\text{x}+3\text{y}-1]^{2}}{13}$
$\Rightarrow13[\text{x}^{2}+\text{y}^{2}-4\text{x}+2\text{y}+5]=4(2\text{x}+3\text{y}-1)^{2}$
$\Rightarrow13\text{x}^{2}+13\text{y}^{2}-52\text{x}+26\text{y}+65=4[2\text{x}+3\text{y}-1]^{2}$
$\Rightarrow13\text{x}^{2}+13\text{y}^{2}-52\text{x}+26\text{y}+65\\=4\Big[(2\text{x)}^{2}+(3\text{y})^{2}+(-1)^{2}+2\times2\text{x}\times3\text{y}\times(-1)+2\times(-1)\times2\text{x}\Big]$
$\Rightarrow13\text{x}^{2}+13\text{y}^{2}-52\text{x}+26\text{y}+65=4\Big[4\text{x}^{2}+9\text{y}^{2}+1+12\text{xy}-6\text{y}-4\text{x}\Big]$
$\Rightarrow13\text{x}^{2}+13\text{y}^{2}-52\text{x}+26\text{y+65}=16\text{x}^{2}+36\text{y}^{2}+4+48\text{xy}-24\text{y}-16\text{x}$
$\Rightarrow16\text{x}^{2}-13{\text{x}^{2}+36}\text{y}^{2}-13\text{y}^{2}+48\text{xy}-16\text{x}+52\text{x}-24\text{y}-26\text{y}+4-65=0$
$\Rightarrow3\text{x}^{2}+23\text{y}^{2}+48+36\text{x}-50\text{y}-61=0$
This is the required equation of the hyperbola.
View full question & answer→Question 295 Marks
Find the equation of the hyperbola whose
foci are (4, 2) and (8,2) and eccentricity is 2.
AnswerThe center of the hyperbola is the mid-point of the line line joining the two foci. So, the coordinates of the centre are $\Big(\frac{4+8}{2},\frac{2+2}{2}\Big)$ i, e.,(6, 2). Let $2\text{a} $ and $2\text{b}$ be the lenght of transverse and conjugate axes and let e be the eccentricity. Then, the equation of the hyperbola is $\frac{(\text{x}-6)^{2}}{\text{a}^{2}}-\frac{(\text{y}-2)^{2}}{\text{b}^{2}}=1$ ----(i) Now, distance between two foci = $2\text{ae}$ $\Rightarrow\sqrt{(8-4)^{2}+(2-2)^{2}}=2\text{ae}$ $\big[$ $\because\text{Foci}=(4, 2)$ and (8, 2) $\big]$ $\Rightarrow\sqrt{(4)^{2}}=2\text{ae}$ $\Rightarrow\text{2ae}=4$ $\big[\because\text{e}=2\big]$ $\Rightarrow2\times\text{a}\times2=4$ $\Rightarrow\text{a}=\frac{4}{4}=1$ $\Rightarrow\text{a}^{2}=1{}$ Now, $\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$ $\Rightarrow\text{b}^{2}=1(2^{2}-1)$ $\big[\because\text{e}=2\big]$ $\Rightarrow\text{b}^{2}=4-1$ $\Rightarrow\text{b}^{2}=3$ Now, $\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$ $\Rightarrow\text{b}^{2}=1(2^{2}-1)$ $\Rightarrow\text{b}^{2}=4-1$ $\text{b}^{2}=3$ Putting $\text{a}^{2}=1$ and $\text{b}^{2}$ = 3 in equation (i), we get $\frac{(\text{x}-6)^{2}}{1}-\frac{(\text{y}-2)^{2}}{3}=1$ $\Rightarrow\frac{3\text{(x}-6)^{2}-(\text{y}-2)^{2}}{3}=1$ $\Rightarrow3\text{(x}-6)^{2}-(\text{y}-2)^{2}=3$ $\Rightarrow3\big[\text{x}^{2}+36-12\text{x}\big]-\big[\text{y}^{2}+4-4\text{y}\big]=3$ $\Rightarrow3\text{x}^{2}+108-36\text{x}-\text{y}^{2}-4+4\text{y}=3$ $\Rightarrow3\text{x}^{2}-\text{y}^{2}-36\text{x}+4\text{y}+101=0$ This is the equation of the requierd hyperbola.
View full question & answer→Question 305 Marks
In each of the following find the equation of the hyperbola satisfying the given conditions
vertices $(0, \pm6)$ $\text{e}=\frac{5}{3}$ [NCERT EXEMPLAR]
AnswerSince, the vertices line on x-axies, so let the equation of the required hyperbola be
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1---(\text{i})$
The lenght of the vertices of the required hyperbola are $(\pm\text{a},0).$
$\therefore\text{a}=7$ [$\because$ vertices = $(\pm7, 0)$]
$\Rightarrow\text{a}^{2}=49---(\text{ii})$
Now,
$\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$
$\Rightarrow\text{b}^{2}=49\Big[(\frac{4}{3})^{2}-1\Big]$ $\big[\because\text{e}=\frac{4}{3}\big]$
$\Rightarrow\text{b}^{2}=49\big[\frac{16}{9}-1\big]$
$\Rightarrow\text{b}^{2}=49\big[\frac{7}{9}\big]$
$\Rightarrow\text{b}^{2}=\frac{343}{9}$
Putting $\text{a}^{2}=49$ and $\text{b}^{2}=\frac{343}{9}$ in equation (i), we get
$\frac{\text{x}^{2}}{49}-\frac{\text{y}^{2}}{\frac{343}{9}}=1$
$\Rightarrow\frac{\text{x}^{2}}{49}-\frac{9\text{y}^{2}}{343}=1$
Hence, the equation of the required hyperbola is
$\frac{\text{x}^{2}}{49}-\frac{9\text{y}^{2}}{343}=1.$
View full question & answer→Question 315 Marks
Find the eccentricity, coordinates of the foci, equation of the directrices and lenght of the latus-rectum of the hyperbola$9\text{x}^{2}-16\text{y}^{2}=-144$
AnswerWe have,
$16\text{x}^{2}-9\text{y}^{2}=-144$
$\Rightarrow\frac{16\text{x}^{2}}{144}-\frac{9\text{y}^{2}}{144}=-1$
$\Rightarrow\frac{\text{x}^{2}}{9}-\frac{\text{y}^{2}}{16}=-1$
This is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=-1,$ Where $\text{a}^{2}-9$ and $\text{b}^{2}-16$
$\therefore\text{a}=3$ and $\text{b}=4$
Eccentricity: The eccentricity e is given by
$\text{e}=\sqrt{1+\frac{\text{a}^{2}}{\text{b}^{2}}}{}$
$=\sqrt{1+\frac{9}{16}}$
$=\sqrt{\frac{25}{16}}$
$=\frac{5}{4}$
Foci: The coordinates of the foci are $(0, \pm\text{be})$.
$\therefore(0, \pm\text{be})=\Big(0, \pm4\times\frac{5}{4}\Big)$
$=(0, \pm5)$
$\therefore$ the coordinates of the foci are $(0, \pm5)$
Equations of the directrices: The equations of the directrices are
$\text{y}=\frac{\pm\text{b}}{\text{e}}$
$\Rightarrow\text{y}=\pm\frac{\frac{4}{5}}{4}-\pm\frac{16}{5}$
$\Rightarrow5\text{y}\mp16-0$
Latus-recutum: The lenght of the latus-rectum
$=\frac{2\text{a}^{2}}{\text{b}}$
$=\frac{2\times9}{4}=\frac{9}{2}$
View full question & answer→Question 325 Marks
focus is $(2, 2),$ directive is $x + y = 9$ and eccentricity $= 2.$
AnswerLet $S (2, 2)$ be the focus and $P(x, y)$ be a point on the hyperbola.
Draw PM perpendicular from $P$ on the directrix. Then, by definition
$sP = ePM$
$\Rightarrow sP^2 = e^2PM^2$
$\Rightarrow(\text{x}-2)^2+(\text{y}-2)^2=2^2\Big[\frac{\text{x}+\text{y}-9}{\sqrt{1^2+1^2}}\Big]$ $\Big[\because\ \text{e}=\frac{4}{3}\Big]$}
$\Rightarrow\ \text{x}^2+4-4\text{x}+\text{y}^2+4-4\text{y}=\frac{4[\text{x+y}-9]^2}{2}$
$\Rightarrow x^2 + y^2 - 4x - 4y + 8 = 2[x + y - 9]^2$
$\Rightarrow x^2 + y^2 - 4x - 4y + 8 = 2[x + y + (-9) + 2 \times x \times y + 2 \times y \times (-9) + 2 \times (-9) \times x]$
$\Rightarrow x^2 + y^2 - 4x - 4y + 8 = 2[x^2 + y^2 + 81 + 2xy - 18y + 18x]$
$\Rightarrow x^2 + y^2 - 4x - 4y + 8 = [2x^2 + 2y^2 + 162 + 4xy - 36y + 36x]$
$\Rightarrow 2x^2 - x^2 + 2y^2 - y^2 + 4xy - 36x + 4x - 36y + 4y + 162 - 8 = 0$
$\Rightarrow x^2 + y^2 - 4xy - 32x - 32y + 154 = 0$
This is the required equation of the hyperbola.
View full question & answer→Question 335 Marks
Find the center eccentricity foci and directrices of the hyperbola
$16\text{x}^{2}-9\text{y}^{2}+32\text{x}+36\text{y}-164=0$
AnswerWe have,
$16\text{x}^{2}-9\text{y}^{2}+32\text{x}+36\text{y}-164=0$
$\Rightarrow16\text{x}^{2}+32\text{x}-9\text{y}^{\text{2}}+36\text{y}-14=0$
$\Rightarrow16(\text{x}^{2}+2\text{x})-9(\text{y}^{2}+4\text{y})-164=0$
$\Rightarrow16[\text{x}^{2}+2\text{x}+1-1]-9[\text{y}^{2}-4\text{y}+4-4]-164=0$
$\Rightarrow16[(\text{x+1})^{2}-1]-9[(\text{y}-2)^{2}-4]-164=0$
$\Rightarrow16(\text{x+1)}^{2}-16-9(\text{y}-2)^{2}+36-164=0$
$\Rightarrow16(\text{x}+1)^{2}-9(\text{y}-2)^{2}+20-164=0$
$\Rightarrow16(\text{x}+1)^{2}-9(\text{y}-2)^{2}-144=0$
$\Rightarrow16(\text{x}+1)^{2}-9(\text{y}-2)^{2}=144$
$\Rightarrow\frac{16(\text{x}+1)^{2}}{144}-\frac{9(\text{y}-2)^{2}}{144}=1$
$\Rightarrow\frac{\text{(x}+1)^{2}}{9}-\frac{(\text{y}-2)}{16}=1$ ---(i)
Shifting the origin at (-1, 2) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by X and Y,
We have,
$\text{x}=\text{x}-1$ and $\text{y}=\text{y}+2$---- (ii)
This is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1$, where $\text{a}^{2}=9$ and $\text{b}^{2}=16.$ so,
We have,
Centre: The coordinates of the centre w.r.e the new axes are $(\text{x}=0, \text{y}=0)$
$\therefore\text{x}=-1$ and $\text{y}=2$ [Using equation (ii)]
So, the coordinates of the centre w.r.t the old axes are (-1, 2)
Eccentricity: The eccentricity e is given by
$\text{e}=\sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$
$=\sqrt{1+\frac{16}{9}}$
$=\sqrt{\frac{25}{9}}$
$=\frac{5}{3}$
Foci: The coordinates of the foci with respect to the new axes are given by $(\text{x}=\pm\text{ae}, \text{y}=0)$
i.e., $(\text{x}=\pm5, \text{y}=0).$
Putting $\text{x}=\pm5$ and $\text{y}=0$ in equation (ii), we get
$\text{x}=\pm5-1$ and $\text{y}=0+2$
$\Rightarrow\text{x}=4,-6$ and $\text{y}=2$
$$ Equation of the directix: The equation of the directrix are
$\text{x}=\pm\frac{\text{a}}{\text{e}}$
$=\pm\frac{\frac{3}{5}}{3}$
$\text{x}=\pm\frac{9}{5}$
Putting $\text{x}=\pm\frac{9}{5}$ in equation (ii), we get
$\text{x}=\pm\frac{9}{5}-1$
$\Rightarrow\text{x}=\frac{\pm9-5}{5}$
$\Rightarrow\text{x}=\frac{4}{5}$ and $\text{x} \frac{-14}{5}$
So the equations of the directrices w.r.t the old axes are
$5\text{x}-4=0$ and $5\text{x}+14=0.$
View full question & answer→Question 345 Marks
Find the center, eccentricity, foci and directrices of the hyperbola
$\text{x}^{2}-\text{y}^{2}+4\text{x}=0$
AnswerWe have, $\text{x}^{2}-\text{y}^{2}+4\text{x}=0$ $\Rightarrow\text{x}^{2}+4\text{x}-\text{y}^{2}=0$ $\Rightarrow\text{x}^{2}+4\text{x}+4-4-\text{y}^{2}=0$ $\Rightarrow(\text{x}+2)^{2}-\text{y}^{2}=4$ $\Rightarrow\frac{(\text{x}+2)^{2}}{4}-\frac{\text{y}^{2}}{4}=1$----(i) Shifting the origin at $(-2, 0)$ without rotating the axes and denoting the new coordinates w.r.t these axes by X and Y, We have, $\text{x}=\text{x}-2$ and $\text{y}=\text{y}$ ----(ii) Using these relations, equation (i) reduces to $\frac{\text{x}^{2}}{4}-\frac{\text{y}^{2}}{4}=1$ This is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1$, where a $\text{a}^{2}=4$ and $\text{b}^{2}=4,$ so. We have, Centre: The coordinates of the centre w.r.t the new axes are $(\text{x}=0, \text{y}=0)$ Putting $\text{x}=0$ and $\text{y}=0$ in equation (ii), we get $\text{x}=-2$ and $\text{y}=0.$ So, the coordinates of the centre w,r,t the old axes are $(-2, 0).$ Eccentricity: The ecentricity e is given by $\text{e}=\sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$ $=\sqrt{1+\frac{4}{4}}$ $=\sqrt{1+1}$ $=\sqrt{2}$ Foci: The coordinates of the foci w.r.t the new axes are $(\text{x}=\pm\text{ae},\text{y}=0)$ i.e, $(\text{x}=\pm2\sqrt{2},\text{y}=0).{}$ Putting $\text{x}=\pm2\sqrt{2}$ and $\text{y}=0$ in equation (ii), we get $\text{x}=\pm2\sqrt{2}-2$ and $\text{y}=0$ $\Rightarrow\text{x}=-2\pm2\sqrt{2}$ and $\text{y}=0$ So, the coordinates of foci w.r.t the old axes are $(-2\sqrt{2}, 0)$ Directrices: The equations of the directrices w.r.t the new axes are Putting $\text{x}=\pm\frac{2}{\sqrt{2}}$ in equation (ii), we get $\text{x}=\pm\frac{2}{\sqrt{2}}-2$ $\Rightarrow\text{x}+2=\pm\frac{\sqrt{2}\times\sqrt{2}}{\sqrt{2}}$ $\Rightarrow\text{x}+2=\pm\sqrt{2}$So,the equations of the directrices w.r.t the old axes $\text{x}+2=\pm\sqrt{2}.$
View full question & answer→Question 355 Marks
Find the eccentricity, coordinates of the foci, equation of the directrices and lenght of the latus-rectum of the hyperbola
$4\text{x}^{2}-3\text{y}^{2}=36$
AnswerWe have,
$4\text{x}^{2}-3\text{y}^{2}=36$
$\Rightarrow\frac{\text{4x}^{2}}{36}-\frac{\text{3y}^{2}}{36}=1$
$\Rightarrow\frac{\text{x}^{2}}{9}-\frac{\text{y}^{2}}{12}=1$
This is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=-1$, Where $\text{a}^{2}=9$ and $\text{b}^{2}=12$
$\therefore\text{a}=3$ and $\text{b}=\sqrt{12}=2\sqrt{3}$
Eccentricity: The eccentricity e is given by
$\text{e}=\sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$
$=\sqrt{1+\frac{12}{9}}$
$=\sqrt{1+\frac{4}{3}}$
$=\sqrt{\frac{7}{3}}$
Foci: The coordinates of the foci are $(\pm\text{ae, 0}).$
$\therefore\pm\text{ae}=\pm3\times\sqrt{\frac{7}{3}}$
$=\pm3\times\frac{\sqrt{7}}{\sqrt{3}}$
$=\pm\sqrt{3}\times\sqrt{7}$
$=\pm\sqrt{21}$
$\therefore(\pm\text{ae, 0})=(\pm\sqrt{21, 0})$
$\therefore$ the coordinates of the foci are $(\pm\sqrt{21, 0})$
Equations of the directrices: The equations of the directrices are
$\text{x}=\frac{\pm\text{a}}{\text{e}}$
$\therefore\text{x}=\pm3\times\frac{\frac{1}{\sqrt{7}}}{\sqrt{3}}$
$=\pm\frac{3\sqrt{3}}{\sqrt{7}}$
$\Rightarrow\sqrt{7\text{x}}\mp3\sqrt{3}=0$
$\therefore$ The equations of the directrices are $\sqrt{7\text{x}}\mp3\sqrt{3}=0$
Latus-rectum: The lenght of the latus-rectum
$=\frac{2\text{b}^{2}}{\text{a}}=\frac{2\times12}{3}=8$
View full question & answer→Question 365 Marks
Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola:
$25\text{x}^{2}- 36\text{y}^{2} = 225$
AnswerWe have,
$25\text{x}^{2}-36\text{y}^{2}=225$
$\Rightarrow\frac{25\text{x}^{2}}{225}-\frac{36^{2}}{225}=1$
$\Rightarrow\frac{\text{x}^{2}}{9}-\frac{4\text{y}^{2}}{25}=1$
$\Rightarrow\frac{\text{x}^{2}}{9}-\frac{\frac{\text{y}^{2}}{25}}{4}=1$
$\Rightarrow\frac{\text{x}^{2}}{(3)^{2}}-\frac{\text{y}^{2}}{\Big(\frac{5}{2}\Big)^{2}}=1$
This is of the form $\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1$, where a $=3$ and $\text{b}=\frac{5}{2}$
Lenght of the transverse axis: The lenght of the transverse axis
$=2\text{a}$
$=2\times3=6$
Lenght of the conjugate axis: The lenght of the conjugate axis is
$2\text{b}=2\times\frac{5}{2}=6$
Eccentricity: The eccentricity e is given by
$\text{e}=\sqrt{1+\frac{\text{b}^{2}}{\text{a}^{2}}}$
$=\sqrt{1+\frac{\frac{25}{4}}{9}}$
$=\sqrt{1+\frac{25}{36}}$
$=\sqrt{\frac{61}{36}}$
$=\sqrt{\frac{61}{6}}$
Lenght of LR$=\frac{2\text{b}^{2}}{\text{a}}=\frac{25}{6}$
Foci $(\pm\frac{61}{2}, 0)$
View full question & answer→Question 375 Marks
In each of the followin find the equation of the hyperbola satisying the given conditions:
foci $(\pm5, 0)$, transverse axis = 8 [NCERT]
AnswerSince, the vertices line on y-axis, So let the equation of the required hyperbola be,
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1---(\text{i})$
The lenght of transverse axis = 8
$\therefore\text{2a}=8$ $[\because $ transverse axis is $\text{2a}]$
$\Rightarrow4\times\text{e}=5$ $[\because\text{a}=4]$
$\Rightarrow\text{e}=\frac{5}{4}$
$\Rightarrow\text{e}^{2}=\frac{25}{16}$
Now,
$\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)$
$=16(\frac{25}{16}-1)$
$=16\times\frac{9}{16}$
$=9$
Putting $\text{a}^{2}=16$ and $\text{b}^{2}=9$ in equation (i), we get
$\frac{\text{x}^{2}}{16}-\frac{\text{y}^{2}}{9}=1$
Hence, the equation of the required hyperbola is
$\frac{\text{x}^{2}}{16}-\frac{\text{y}^{2}}{9}=1$
View full question & answer→Question 385 Marks
Find the equation of the hyperbola whosefocus is (1, 1), directrix is 3 x + 4 y + 8 = 0 and eccentricity=2
AnswerLet (1, 1) be the focus and P (x, y) be a point a on the hyperbola, Draw PM perpendicular from P on the directrix, then, by definition
$\text{sP}=\text{ePM}$
$\Rightarrow\text{sPM}^{2}=\text{e}^{2}\text{PM}^{2}$
$\Rightarrow(\text{x}-1)^{2}+(\text{y}-1)^{2}=2^{2}\Bigg[\frac{3\text{x}+4\text{y}+8}{\sqrt{3^{2}+4^{2}}}\Bigg]^{2}$
$\Rightarrow\text{x}^{2}+1-2\text{x}+\text{y}^{2}+1-2\text{y}=4\Bigg[\frac{3\text{x}+4\text{y}+8}{\sqrt{25}}\Bigg]$
$\Rightarrow\text{x}^{2}+\text{y}^{2}-2\text{x}-\text{2y}+2=\frac{4(\text{3x}+4\text{y}+8)^{2}}{25}$
$\Rightarrow25\text{x}^{2}+25\text{y}^{2}-50\text{x}-50\text{y}+50=4(3\text{x}+4\text{y}+8)^{2}$
$\Rightarrow25\text{x}^{2}+25\text{y}^{2}-50\text{x}-50\text{y}+50=4\Bigg[9\text{x}^{2}+16\text{y}^{2}+6\text{y}+24\text{xy}+64\text{y}+48\text{x}\Bigg]$
$\Rightarrow25\text{x}^{2}+25\text{y}^{2}-50\text{x}-50\text{y}+50=36\text{x}^{2}+64\text{y}^{2}+256+96\text{xy}+256\text{y}+192\text{x}$
$\Rightarrow36\text{x}^{2}-25\text{x}^{2}+64\text{y}^{2}-25\text{y}^{2}+96\text{xy}+192\text{x}+50\text{x}+256\text{y}+50\text{y}+256-50=0$
$\Rightarrow11\text{x}^{2}+39\text{y}^{2}+96\text{xy}+242\text{x}+306\text{y}+206=0$
This is the requierd equation of the hyperbola.
View full question & answer→Question 395 Marks
In each of the following find the equation of the hyperbola satisfying the given conditionsfoci $(\pm0,\pm\sqrt10),$ passing throught (2,3) [NCERT ] $$
AnswerSince, the vertices line on x-axies, so let the equation of the required hyperbola be
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1---(\text{i})$
It passes throught (2,3)
$\therefore\frac{(2)^{2}}{\text{a}^{2}}-\frac{(3)^{2}}{\text{b}^{2}}=-1$
$\Rightarrow\frac{4}{\text{a}^{2}}-\frac{9}{\text{b}^{2}}=-1$
$\Rightarrow\frac{4}{\text{a}^{2}}-\frac{9}{\text{a}^{2}(\text{e}^{2}-1)}$ $\big[\because\text{b}^{2}=\text{a}^{2}(\text{e}^{2}-1)\big]$
$\Rightarrow\frac{4}{\text{a}^{2}}-\frac{9}{\text{a}^{2}\text{e}^{2}-\text{a}^{2}}=-1---(\text{ii})$
The coordinates of the required hyperbola are $(0,\pm\text{ae})$
$\therefore\text{ae}=\sqrt{10}$
$\Rightarrow\text{a}^{2}{\text{e}^{2}}=10---\text{(iii)}$
View full question & answer→Question 405 Marks
Show that the set of all points such that the difference of their distances from (4, 0) and (-4, 0) is always equal to 2 represents a hyperbola
AnswerLet the point be (x, y)
$\therefore\Big[\sqrt{(\text{x}-4)^{2}+(\text{y}-0)^{2}}\Big]-\Big[\sqrt{(\text{x}+4)^{2}+(\text{y}-0)^{2}}\Big]=2$
$\Rightarrow\Big[\sqrt{(\text{x}-4)^{2}+(\text{y}-0)^{2}}\Big]^{2}=\Big[2+\sqrt{(\text{x}+4)^{4}+(\text{y}-0)^{2}}\Big]^{2}$
$\Rightarrow(\text{x}-4)^{2}+\text{y}^{2}=4+(\text{x}+4)^{2}+\text{y}^{2}+4\sqrt{(\text{x}+4)^{2}+(\text{y}-0)^{2}}$
$\Rightarrow(\text{x}-4)^{2}-(\text{x}+4)^{2}=4+4\sqrt{(\text{x}+4)^{2}+(\text{y}-0)^{2}}$
$\Rightarrow-16\text{x}=4+4\sqrt{(\text{x}+4)^{2}+(\text{y}-0)^{2}}$
$\Rightarrow-16\text{x}-4=4\sqrt{(\text{x}+4)^{2}+(\text{y}-0)^{2}}$
$\Rightarrow-4(4\text{x}+1)=4\sqrt{(\text{x}-4)^{2}+(\text{y}-0)^{2}}$
$\Rightarrow-4(4\text{x}+1)=\sqrt{(\text{x}-4)^{2}+(\text{y}-0)^{2}}$
$\Rightarrow16\text{x}^{2}+8\text{x}+1=\text{x}^{2}+8\text{x}+16+\text{y}^{2}$
$\Rightarrow15\text{x}^{2}-\text{y}^{2}=15$
$\Rightarrow\frac{\text{x}^{2}}{1}-\frac{\text{y}^{2}}{15}=1$
Which is the equation of a hyperbola.
View full question & answer→Question 415 Marks
In each of the followin find the equation of the hyperbola satisying the given conditions:
vertices $(0, \pm5), $ foci $(0,\pm8)$ [NCERT]
AnswerSince, the vertices line on y-axis, So let the equation of the required hyperbola be
$\frac{\text{x}^{2}}{\text{a}^{2}}-\frac{\text{y}^{2}}{\text{b}^{2}}=1---(\text{i})$
The coordinates of its vertices and foci are $(0, \pm\text{b})$ and $(0, \pm\text{be})$ respectively.
$\therefore\text{b} = 5$ [$$$\because$ vertices = $(0, \pm5)$]
$\Rightarrow\text{b}^{2}-5$
and, $\text{be} = 8$ [$\because$ foci = $(0, \pm8)$]
$\Rightarrow5\times\text{e}=8$ [$\because\text{b}=5$]
$\Rightarrow\text{e}=\frac{8}{5}$
$\Rightarrow\text{e}^{2}=\frac{64}{25}$
Now,
$\text{a}^{2}=\text{b}^{2}(\text{e}^{2}-1)$
$\Rightarrow\text{a}^{2}=25\big(\frac{64}{25}-1\big)$[$\because\text{b}^{2}=25$ and $\text{e}^{2}=\frac{64}{25}$]
$\Rightarrow\text{a}^{2}-25\times\frac{39}{25}$
$\Rightarrow\text{a}^{2}-39$
Putting $\text{a}^{2}-39$ and $\text{b}^{2}-25$ in equatoin (i), we get
$\frac{\text{x}^{2}}{39}-\frac{\text{y}^{2}}{25}=-1$
Hence, the equation of the required hyperbola is
$\frac{\text{x}^{2}}{39}-\frac{\text{y}^{2}}{25}=-1$
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