Question
A hyperbola having the transverse axis of length $$2\,\sin \,\theta ,$$ is confocal with the ellipse $$3{x^2} + 4{y^2} = 12.$$ Then its equation is :
A.
$${x^2}{\text{cose}}{{\text{c}}^2}\theta - {y^2}{\sec ^2}\theta = 1$$
B.
$${x^2}{\sec ^2}\theta - {y^2}{\text{cose}}{{\text{c}}^2}\theta = 1$$
C.
$${x^2}{\sin ^2}\theta - {y^2}{\cos ^2}\theta = 1$$
D.
$${x^2}{\cos ^2}\theta - {y^2}{\sin ^2}\theta = 1$$
Answer :
$${x^2}{\text{cose}}{{\text{c}}^2}\theta - {y^2}{\sec ^2}\theta = 1$$
Solution :
Equation of the ellipse is $$3{x^2} + 4{y^2} = 12$$
$$ \Rightarrow \frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1......\left( 1 \right)$$
Eccentricity $${e_1} = \sqrt {1 - \frac{3}{4}} = \frac{1}{2}$$
So, the foci of ellipse are $$\left( {1,\,0} \right)$$ and $$\left( { - 1,\,0} \right)$$
Let the equation of the required hyperbola be
$$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1......\left( 2 \right)$$
Given $$2a = 2\,\sin \,\theta \Rightarrow a = \sin \,\theta $$
Since the ellipse $$\left( 1 \right)$$ and the hyperbola $$\left( 2 \right)$$ are confocal, so the foci of hyperbola are $$\left( {1,\,0} \right)$$ and $$\left( { - 1,\,0} \right)$$ too.
If the eccentricity, of hyperbola be $${e_2}$$ then
$$\eqalign{
& a{e_2} = 1 \Rightarrow \sin \,\theta {e_2} = 1 \Rightarrow {e_2} = {\text{cosec}}\,\theta \cr
& \therefore \,{b^2} = {a^2}\left( {e_2^2 - 1} \right) = {\sin ^2}\theta \left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right) = {\cos ^2}\theta \cr} $$
$$\therefore $$ Required equation of the hyperbola is
$$\frac{{{x^2}}}{{{{\sin }^2}\theta }} - \frac{{{y^2}}}{{{{\cos }^2}\theta }} = 1 \Rightarrow {x^2}{\text{cose}}{{\text{c}}^2}\theta - {y^2}{\sec ^2}\theta = 1$$