Hyperbola (S.N.Dey) | Part-5

In the previous article , we have solved few Short Answer Type questions of Hyperbola chapter of S.N.Dey mathematics, Class 11. In this article , we will solve few more Short Answer Type questions of Hyperbola related problems of S N Dey mathematics class 11.

6. Find the equation of the hyperbola, referred to its axes as lines parallel to co-ordinate axes and having centre at $~(3,-2),$ eccentricity $~\frac{\sqrt{5}}{2}$ and the length of latus rectum $~2$ .

Solution.

The equation of the hyperbola having centre at $~(3,-2)~$ and axes parallel to co-ordinate axes can be written as

$~\frac{(x-3)^2}{a^2}-\frac{(y+2)^2}{b^2}=1 \rightarrow(1)$

The length of latus rectum $~\frac{2b^2}{a}=2 \Rightarrow b^2=a$

$e=\frac{\sqrt{5}}{2}~~\text{(given)} \\ \text{or,}~~ e^2=\frac 54 \\ \text{or,}~~ 1+\frac{b^2}{a^2}=\frac 54 \\ \text{or,}~~ \frac{a}{a^2}=\frac 54-1 \\ \text{or,}~~ \frac 1a=\frac 14 \\ \text{or,}~~ a=4.$

$\therefore~ b^2=a=4$

Hence, by $(1)$ the required equation of hyperbola is given by

$~\frac{(x-3)^2}{16}-\frac{(y+2)^2}{4}=1 \Rightarrow (x-3)^2-4(y+2)^2=16.$

7. The distance between the vertices of a hyperbola is half the distance between its foci and the length of its semi-conjugate axis is $~6$ . Referred to its axes as axes of co-ordinates, find the equation of the hyperbola.

Solution.

We know that the distance between the vertices of the hyperbola is $~2a~$ unit and the distance between the foci is $2ae$ unit.

By question, $~2a=\frac{2ae}{2} \Rightarrow e=2.$

Again, the length of semi-conjugate axis $(b)=6.$

Now, $b^2=a^2(e^2-1) \Rightarrow 36=a^2(2^2-1) \Rightarrow a^2=\frac{36}{3}=12.$

Hence, the equation of the required hyperbola is

$~\frac{x^2}{12}-\frac{y^2}{36}=1 \Rightarrow 3x^2-y^2=36.$

8. The co-ordinates of the foci of a hyperbola are $~(16, 0)~$ and $~(-8,0)~$ and its eccentricity is $~3;$ find the equation of the hyperbola and the length of its latus rectum.

Solution.

$2ae=\text{the distance between the foci of the hyperbola} \\ \text{or,}~~ 2ae=\sqrt{(16+8)^2+(0-0)^2} \\ \text{or,}~~ 2ae=\sqrt{(24)^2}=24 \\ \text{or,}~~ ae=\frac{24}{2}=12 \\ \text{or,}~~ a \times 3=12 \\ \text{or,}~~ a=\frac{12}{3}=4.$

$~b^2=a^2(e^2-1) =4^2(3^2-1) \\ \text{or,}~~b^2=16 \times 8=128.$

Now, the centre of the hyperbola is the mid-point of the foci i.e., $~\left(\frac{16-8}{2},\frac{0+0}{2}\right)=(4,0).$

From the co-ordinates of the foci, we can notice that the transverse axis of the hyperbola is along $~x$ -axis and so the equation of the hyperbola having centre at $(4,0)$ is

$~\frac{(x-4)^2}{16}-\frac{(y-0)^2}{128}=1 \Rightarrow 8(x-4)^2-y^2=128.$

Computer Science with Python Textbook for Class 11 Paperback

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9. The centre of a hyperbola is at $~(2, 4)~$ , the co-ordinates of one of its vertices are $~(6, 4)~$ and eccentricity is $~\sqrt{5};$ find its equation.

Solution.

If $a$ is the distance of the vertex from the centre , then

$a=\sqrt{(6-2)^2+(4-4)^2}=\sqrt{4^2+0^2}=4$

Again, $~b^2=a^2(e^2-1)=4^2[(\sqrt{5})^2-1]=16 \times 4=64$

Since the ordinates of the centre and the vertex of the hyperbola is $4$ , both cantre and the vertex lie on the straight line $~y=4.$

So, the transverse axis of the hyperbola is parallel to $x$ -axis. So, the equation of the hyperbola with centre $~(\alpha,\beta)~$ is

$~\frac{(x-\alpha)^2}{a^2}-\frac{(y-\beta)^2}{b^2}=1 \\ \text{or,}~~ \frac{(x-2)^2}{16}-\frac{(y-4)^2}{64}=1 \\ \therefore~ 4(x-2)^2-(y-4)^2=64.$

10. Find the equation of the hyperbola with centre at the origin, transverse axis on $~x$ -axis, passing through the point $~\left(3\sqrt{2},-2\right)~$ and having the length of semi-conjugate axis $~2.$

Solution.

The hyperbola having centre at the origin and transverse axis on $~x$ -axis, can be written in the form of $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\rightarrow(1)$

Since the hyperbola $(1)$ passes through the point $~(3\sqrt{2},-2),~$

$\therefore~\frac{(3\sqrt{2})^2}{a^2}-\frac{(-2)^2}{b^2}=1 \\ \text{or,}~\frac{18}{a^2}-\frac{4}{2^2}=1~~[~\because~ b=2~] \\ \text{or,}~~ \frac{18}{a^2}-1=1 \\ \text{or,}~~\frac{18}{a^2}=1+1 \\ \text{or,}~~a^2=\frac{18}{2}=9 .$

Hence, putting the values of $a$ and $b$ in $(1)$ , we get the required equation of hyperbola which is given by

$~\frac{x^2}{9}-\frac{y^2}{4}=1 \Rightarrow 4x^2-9y^2=36.$

11. The lengths of transverse and conjugate axes of a hyperbola are $~4~$ unit and $~6~$ unit respectively and their equations are $~x+3=0~$ and $~y-1=0~;$ find the equation of the hyperbola.

Solution.

It is given that $~~2a=4 \Rightarrow a=2,~2b=6\Rightarrow b=3.$

Since the transverse and conjugate axes of the hyperbola are $~x+3=0 \rightarrow(1)~$ and $~ y-1=0 \rightarrow(2)~$ , the point of intersection of $~(1)~$ and $~(2)~$ is given by $~(-3,1)~$ which is the centre of the hyperbola.

Since the transverse axis of the hyperbola given by $~x+3=0~$ is parallel to $~y$ -axis, so the equation of the hyperbola with centre $(\alpha,\beta)$ is given by

$~\frac{(y-\beta)^2}{a^2}-\frac{(x-\alpha)^2}{b^2}=1 \\ \text{or,}~~ \frac{(y-1)^2}{2^2}-\frac{(x+3)^2}{3^2}=1 \\ \text{or,}~~ \frac{(y-1)^2}{4}-\frac{(x+3)^2}{9}=1 \\ \therefore~ 9(y-1)^2-4(x+3)^2=36.$

12. $~e_1~$ and $~e_2~$ are respectively the eccentricities of a hyperbola and its conjugate. Prove that, $~\frac{1}{e_1^2}+\frac{1}{e_2^2}=1.$

Solution.

Let the equation of the hyperbola be $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \rightarrow(1)$

The equation of the conjugate hyperbola of $(1)$ can be written as $~\frac{y^2}{b^2}-\frac{x^2}{a^2}=1 \rightarrow(2)$

$\therefore~ e_1=\sqrt{1+\frac{b^2}{a^2}},~~e_2=\sqrt{1+\frac{a^2}{b^2}}.$

$\therefore~ \frac{1}{e_1^2}+\frac{1}{e_2^2}=\frac{1}{1+\frac{b^2}{a^2}}+\frac{1}{1+\frac{a^2}{b^2}}=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=\frac{a^2+b^2}{a^2+b^2}=1~~\text{(proved)}$

13. Find the eccentricity of the hyperbola $~x^2-y^2=1.$ If $~S, S_1~$ are the foci and $~P~$ any point on this hyperbola, prove that, $~CP^2=SP \cdot S_1P$ ( $~C$ is the origin).

Solution.

$x^2-y^2=1 \Rightarrow \frac{x^2}{1^2}-\frac{y^2}{1^2}=1\rightarrow(1)$

Comparing $(1)$ with the general equation of hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ we get $~a=1,~b=1.$

$\therefore~ e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1^2}{1^2}}=\sqrt{1+1}=\sqrt{2}.$

The co-ordinates of foci of the hyperbola $(1)$ are $~( \pm ae,0)~$ i.e., $~(\pm 1 \cdot \sqrt{2},0)=(\pm \sqrt{2},0).$

$\therefore~ S \equiv(\sqrt{2},0),~~S_1 \equiv (-\sqrt{2},0).$

The co-ordinates of any point $P$ on the hyperbola $(1)$ can be taken as $~(\sec\theta,\tan\theta).$

$\therefore~ CP^2=(\sec\theta-0)^2+(\tan\theta-0)^2=\sec^2\theta+\tan^2\theta\rightarrow(2)$

Let $~\sec\theta=c,~~\tan\theta=d~$ so that $~\sec^2\theta=1+\tan^2\theta \Rightarrow c^2=1+d^2\rightarrow(3)$

$\therefore~~P \equiv(c,d).$

Now, $~\overline{SP}^2=(c-\sqrt{2})^2+d^2,~~\overline{S_1P}^2=(c+\sqrt{2})^2+d^2.$

$\therefore~ \overline{SP}^2 \cdot \overline{S_1P}^2\\=[(c-\sqrt{2})^2+d^2] \cdot [(c+\sqrt{2})^2+d^2]\\=(c-\sqrt{2})^2(c+\sqrt{2})^2+d^2[(c+\sqrt{2})^2+(c-\sqrt{2})^2]+d^4\\=(c^2-2)^2+d^2[2(c^2+2)]+d^4\\=c^4-4c^2+4+2d^2(c^2+2)+d^4\\=c^4+2c^2d^2+d^4-4(c^2-1)+4d^2\\=(c^2+d^2)^2-4d^2+4d^2~~[\text{By (3)}]\\=(c^2+d^2)^2 \\ \text{or,}~~ \overline{SP}\cdot \overline{S_1P}=(c^2+d^2)=\sec^2\theta+\tan^2\theta \rightarrow(4)$

Hence by $~(3)~$ and $~(4)~$ , the result follows.

14. $~P~$ is a variable point on the hyperbola $~x^2-y^2=16$ and $~A (8, 0)~$ is a fixed point. Show that the locus of the middle point of the line segment $~\overline{AP}~$ is another rectangular hyperbola.

Solution.

$x^2-y^2=16 \Rightarrow \left(\frac x4\right)^2-\left(\frac y4\right)^2=1\rightarrow(1)$

The co-ordinates of any point $P$ on the hyperbola $(1)$ can be written as $~P(4\sec\theta,4\tan\theta).$

Let the middle point of $AP$ be $~(h,k).$

$\therefore~ h=\frac 12(8+4\sec\theta),~~k=\frac 12(0+4\tan\theta) \\ \Rightarrow h=4+2\sec\theta\rightarrow(2), ~~k=2\tan\theta\rightarrow(3)$

From $(2)$ and $(3)$ we get,

$~\sec\theta=\frac{h-4}{2},~~ \tan\theta=\frac k2.$

$\because~~\sec^2\theta-\tan^2\theta=1 \\ \text{or,}~~ \frac14(h-4)^2-\frac 14k^2=1\rightarrow(4)$

So, by $(4)$ we can say that the locus of the middle point of the line segment $~\overline{AP}~$ is

$~\frac{(x-4)^2}{4}-\frac{y^2}{4}=1 \Rightarrow (x-4)^2-y^2=4\rightarrow(5)$

Since the equation $(5)$ is a rectangular hyperbola, so the result follows.

15. Show that the difference of focal distances of any point of the hyperbola $~3x^2-4y^2 = 48~$ is constant.

Solution.

$3x^2-4y^2=48 \Rightarrow \frac{x^2}{16}-\frac{y^2}{12}=1\rightarrow(1)$

Comparing $(1)$ with the standard form of the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ , we get

$a^2=16 \Rightarrow a=4,~~b^2=12 \Rightarrow b=2\sqrt{3} \\ \therefore~~ e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{12}{16}}=\sqrt{\frac{28}{16}}=\sqrt{\frac 74}=\frac{\sqrt{7}}{2}.$

Now, the co-ordinates of foci are given by

$~(\pm ae,0)=\left( \pm 4 \times \frac{\sqrt{7}}{2},0\right)=(\pm 2\sqrt{7},0).$

Let any point on the hyperbola $(1)$ be $~P(4\sec\theta,2\sqrt{3}\tan\theta).$

$\text{If}~S \equiv(2\sqrt{7},0),~~S'\equiv(-2\sqrt{7},0), \\ \therefore~ SP^2=(4\sec\theta-2\sqrt{7})^2+(2\sqrt{3}\tan\theta-0)^2 \\ \text{or,}~~ SP^2=16\sec^2\theta-16\sqrt{7}\sec\theta+28+12\tan^2\theta \\ \text{or,}~~ SP^2=16\sec^2\theta-16\sqrt{7}\sec\theta+28+12(\sec^2\theta-1) \\ \text{or,}~~ SP^2=28\sec^2\theta-16\sqrt{7}\sec\theta+16 \\ \therefore SP^2=(2\sqrt{7} \sec\theta)^2-2 \cdot 2\sqrt{7} \sec\theta \cdot 4+4^2 \\ \text{or,}~~ SP=2\sqrt{7} \sec\theta-4$

Similarly, we can easily show that

$S'P^2=(4\sec\theta+2\sqrt{7})^2+(2\sqrt{3} \tan\theta)^2=(2\sqrt{7} \sec\theta+4)^2 \\ \text{or,}~~ S'P=2\sqrt{7}\sec\theta+4$

$\therefore~|SP-S'P|\\=|(2\sqrt{7} \sec\theta-4)-(2\sqrt{7} \sec\theta+4)|\\=|2\sqrt{7}\sec\theta-4-2\sqrt{7}\sec\theta-4|\\=|-8|=8$

Hence, the difference of focal distances of any point of the given hyperbola is constant.

16. The co-ordinates of the foci of a hyperbola are $~(0, \pm 4)~$ and the length of its latus rectum is $~12~$ unit; find its equation.

Solution.

By question, we notice that the co-ordinates of the foci $~(0, \pm 4)~$ lie on the $~y$ -axis. Now, the transverse axis of the hyperbola is along $~y$ -axis.

$\therefore~$ The equation of the hyperbola can be written as $~\frac{y^2}{a^2}-\frac{x^2}{b^2}=1 \rightarrow(1)$

Now, $~\frac{2b^2}{a}=12 \Rightarrow b^2=6a \rightarrow(2)$

$~2ae=\sqrt{(0-0)^2+(4+4)^2}=8 \\ \text{or,}~~ ae=\frac 82=4 \\ \text{or,}~~ a^2e^2=4^2 \\ \text{or,}~~ a^2 \left(1+\frac{b^2}{a^2}\right)=16 \\ \text{or,}~~ a^2+b^2=16 \\ \text{or,}~~ a^2+6a=16~~[\text{By (2)}] \\ \text{or,}~~ a^2+6a-16=0 \\ \text{or,}~~ a^2+8a-2a-16=0 \\ \text{or,}~~ a(a+8)-2(a+8)=0 \\ \text{or,}~~(a+8)(a-2)=0 \\ \text{or,}~~ a=2~~[~\because a>0~]$

$\therefore~ b^2=6a=6 \times 2=12.$

So, the equation of the hyperbola is given by

$~\frac{y^2}{4}-\frac{x^2}{12}=1 \Rightarrow 3y^2-x^2=12.$

The eccentricity of the hyperbola $(1)$ is

$~e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{12}{4}}=\sqrt{1+3}=\sqrt{4}=2.$

17. Show that for all values of $t$ the point $~ x=a \cdot \frac{1+t^2}{1-t^2},~y=\frac{2at}{1-t^2}~$ lies on a fixed hyperbola. What is the value of the eccentricity of the hyperbola.

Solution.

$~x=a \cdot \frac{1+t^2}{1-t^2},~~y=\frac{2at}{1-t^2} \\ \Rightarrow \frac xa=\frac{1+t^2}{1-t^2},~~\frac ya=\frac{2t}{1-t^2}.$

$\therefore~ \frac{x^2}{a^2}-\frac{y^2}{a^2}=\left(\frac{1+t^2}{1-t^2}\right)^2-\left(\frac{2at}{1-t^2}\right)^2 \\ \text{or,}~~ \frac{x^2}{a^2}-\frac{y^2}{a^2}=\frac{(1+t^2)^2-4t^2}{(1-t^2)^2} \\ \text{or,}~~ \frac{x^2}{a^2}-\frac{y^2}{a^2}=\frac{(1-t^2)^2}{(1-t^2)^2} \\ \text{or,}~~ \frac{x^2}{a^2}-\frac{y^2}{a^2}=1\rightarrow(1)$

Clearly, the equation $(1)$ represents a hyperbola whose eccentricity is given by $~e=\sqrt{1+\frac{a^2}{a^2}}=\sqrt{1+1}=\sqrt{2}.$

Mathematics by S.N.DE for class 12 (WBHS,WBJEE, JEE- Main, JEE- Advanced) Paperback

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Hyperbola (S.N.Dey) | Part-5 | Ex-6

In the previous article , we have solved few Short Answer Type questions of Hyperbola chapter of S.N.Dey mathematics, Class 11. In this article , we will solve few more Short Answer Type questions of Hyperbola related problems of S N Dey mathematics class 11.

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