Solve the Following Question.(4 Marks) - Maths STD 11 Science Questions - Vidyadip

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Solve the Following Question.(4 Marks)

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16 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

Find the equation of the hyperbola whose
Vertices are (-8, -1) and (-4, 4) and focus is (17, -1)

Answer

The 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.

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Question 24 Marks

If P is any point on the hyperbola whose axis are equal, prove that SP . SP =$\text{CP}^{2}$

Answer

For 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 $\text{SP}^{2}.\text{SP}^{2}$ $=(2\alpha^{2}+\alpha^{2}+\beta^{2})^{2}-8\text{a}^{2}\alpha^{2}$
$=4\alpha^{4}+4\alpha^{2}(\alpha^{2}+\beta^{2})+(\alpha^{2}+\beta^{2})-8\text{a}^{2}\text{a}^{2}$
$=4\text{a}^{2}(\alpha^{2}-2\alpha^{2})+4\text{a}^{2}(\alpha^{2}+\beta^{2})+(\alpha^{2}+\beta^{2})^{2}$
$=4\alpha^{2}(\alpha^{2}-\beta^{2}-2\alpha^{2})+4\text{a}^{2}(\alpha^{2}+\beta^{2})+(\alpha^{2+}\beta^{2})^{2}$
$=(\alpha^{2}+\beta^{2})^{2}=\text{CP}^{4}$
$\text{SP. }\text{SP}=\text{CP}^{2}$

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Question 34 Marks

Find the equation of the hyperbola whoseFoci are (6, 4) and (-4, 4) and eccentricity is 2.

Answer

The 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.

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Question 44 Marks

Find the center, eccentricity, foci and directrices of the hyperbola
$\text{x}^{2}-3\text{y}^{2}-2\text{x}=8.$

Answer

We 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
$\text{x}-\pm\frac{9}{2\sqrt{3}}+1$
$\Rightarrow\text{x}=\pm\frac{9}{2\sqrt{3}}$

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Question 54 Marks

Find the eccentricity of the hyperbola, the length of whose conjugate axis is $\frac{3}{4}$ of the length of transvers axis.

Answer

Let $\text{2a}$ 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 $=\frac{5}{4}$

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Question 64 Marks

Find the equation of the hyperbola whose Vertices are at $(0 \pm 7)$ and foci at $\big(0, + \frac{28}{3}\big).$

Answer

Since, 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.

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Question 74 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]

Answer

Since, 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
$\frac{\text{x}^{2}}{25}-\frac{\text{y}^{2}}{20}=1$.

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Question 84 Marks

In each of the followin find the equation of the hyperbola satisying the given conditions:

vertices $(0, \pm3)$, foci $(0, \pm5)$ [NCERT]

Answer

Since, 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

$\frac{\text{x}^{2}}{16}-\frac{\text{y}^{2}}{9}=-1$

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Question 94 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 $=\sqrt{2}$

Answer

Let 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 $\text{x}^{2}-\text{y}^{2}=32.$

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Question 104 Marks

If the distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$, then obtain its equation.

Answer

Eccentricity = $\text{e}=\sqrt{2}$ 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, $\text{x}^{2}-\text{y}^{2}=32$

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Question 114 Marks

In each of the following find the equations of the hyperbola satisfying the given conditions:Vertices $(\pm2, 0)$, foci $(\pm3, 0)$ [NCERT]

Answer

Let 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 $\frac{\text{x}^{2}}{4}-\frac{\text{y}^{2}}{5}-1$.

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Question 124 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

Answer

Let 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 $25\text{x}^{2}-144\text{y}^{2}=900.$

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Question 134 Marks

In each of the following find the equation of the hyperbola satisfying the given conditions
foci $(0, \pm12), $ latus-rectum=36 [NCERT]

Answer

Since, 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 $3\text{y}^{2}-\text{x}^{2}=108$

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Question 144 Marks

Find the equation of the hyperbola whoseFoci at $(\pm2, 0)$ and eccentricity is $\frac{3}{2}$. [NCRT EXEMPLAR]

Answer

The 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
$\frac{9\text{x}^{2}}{16}-\frac{9\text{y}^{2}}{20}=1$
$\Rightarrow\frac{\text{x}^{2}}{4}-\frac{\text{y}^{2}}{5}=\frac{4}{9}$

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Question 154 Marks

In each the following find the equation of the hyperbola satisfying the given conditions:
Foci $(\pm4, 0), $ latus-rectum = 12 [NCERT]

Answer

Since, 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
$\frac{\text{x}^{2}}{4}-\frac{\text{y}^{2}}{12}=1.$

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Question 164 Marks

Find the equation of the hyperbola whose
Focus is at $(4, 2)$ centre at $(6, 2)$ and $e = 2.$

Answer

The 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 = $(\text{ae}+\text{x}_0, \text{y}_0)$
$\therefore\text{ae} = -2$
$\Rightarrow\text{a} = -1$
$\text{b}^{2} = (-2)^{2}-\text{a}^{2}$
$\Rightarrow\text{b}^{2}=(-2)^{2}-(-1)^{2}$
$\Rightarrow\text{b}^{2}=3$
$\Rightarrow\frac{\text{(x}-6)^{2}}{1}-\frac{(\text{y}-2)^{2}}{3}=1$
$\Rightarrow3\text{(x}-6)-(\text{y}-2)^{2}=3$

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