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.
Hyperbola-S N Dey Solutions
Hyperbola-Short Answer Type Questions and Solutions :

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}.


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