Question 15 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 25 Marks
Find the equation of the hyperbola whose Foci 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 35 Marks
Find the equation of the hyperbola whose
Foci 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 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 55 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 65 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 75 Marks
Find the equation of the hyperbola whose Focus is at (5, 2), vertex at (4, 2) and center at (3, 2)
AnswerThe equation of the hyperbola with center (X0, Y0) 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 85 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 95 Marks
In each of the following find the equation of the hyperbola satisfying the given conditions foci $(\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 105 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 115 Marks
Find the equation of the hyperbola whose focus 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 125 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 135 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 145 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 155 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 165 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 175 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 185 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 195 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 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 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 225 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 235 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 245 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 255 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 265 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 275 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 285 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
⇒ sP2 = e2PM2
$\Rightarrow(\text{x}-2)^2+(\text{y}-2)^2=2^2\Big[\frac{\text{x}+\text{y}-9}{\sqrt{1^2+1^2}}\Big]$ $\Big[\because\ \text{e}=\frac{4}{3}\Big]$
$\Rightarrow\ \text{x}^2+4-4\text{x}+\text{y}^2+4-4\text{y}=\frac{4[\text{x+y}-9]^2}{2}$
⇒ x2 + y2 - 4x - 4y + 8 = 2[x + y - 9]2
⇒ x2 + y2 - 4x - 4y + 8 = 2[x + y + (-9) + 2 × x × y + 2 × y × (-9) + 2 × (-9) × x]
⇒ x2 + y2 - 4x - 4y + 8 = 2[x2 + y2 + 81 + 2xy - 18y + 18x]
⇒ x2 + y2 - 4x - 4y + 8 = [2x2 + 2y2 + 162 + 4xy - 36y + 36x]
⇒ 2x2 - x2 + 2y2 - y2 + 4xy - 36x + 4x - 36y + 4y + 162 - 8 = 0
⇒ x2 + y2 - 4xy - 32x - 32y + 154 = 0
This is the required equation of the hyperbola.
View full question & answer→Question 295 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 305 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 315 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 325 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 335 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 345 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
$\frac{\text{x}^{2}}{39}-\frac{\text{y}^{2}}{25}=-1$
View full question & answer→Question 355 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 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
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 385 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 395 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 405 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 415 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 (X0, Y0) is given by
$\frac{\text{(x}-\text{x}_0)}{\text{a}^{2}}-\frac{(\text{y}-\text{y}_0)^{2}}{\text{b}^{2}}=1$
Focus = $(\text{ae}+\text{x}_0, \text{y}_0)$
$\therefore\text{ae} = -2$
$\Rightarrow\text{a} = -1$
$\text{b}^{2} = (-2)^{2}-\text{a}^{2}$
$\Rightarrow\text{b}^{2}=(-2)^{2}-(-1)^{2}$
$\Rightarrow\text{b}^{2}=3$
$\Rightarrow\frac{\text{(x}-6)^{2}}{1}-\frac{(\text{y}-2)^{2}}{3}=1$
$\Rightarrow3\text{(x}-6)-(\text{y}-2)^{2}=3$
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