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 :
The length of transverse axis $$ = 2\,\sin \,\theta = 2a\,\, \Rightarrow a = \sin \,\theta $$
Also for ellipse $$3{x^2} + 4{y^2} = 12$$
$$\eqalign{
& {\text{or}}\,\,\,\,\,\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1,\,\,{a^2} = 4,\,\,{b^2} = 3 \cr
& e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 - \frac{3}{4}} = \frac{1}{2} \cr} $$
$$\therefore $$ Focus of ellipse $$ = \left( {2 \times \frac{1}{2},\,0} \right) \Rightarrow \left( {1,\,0} \right)$$
As hyperbola is confocal with ellipse, focus of hyperbola $$ = \left( {1,\,0} \right) \Rightarrow ae = 1 \Rightarrow \sin \,\theta \times e = 1 \Rightarrow e = {\text{cosec}}\,\theta $$
$$\therefore {b^2} = {a^2}\left( {{e^2} - 1} \right) = {\sin ^2}\theta \left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right) = {\cos ^2}\theta $$
$$\therefore $$ Equation of hyperbola is
$$\eqalign{
& \frac{{{x^2}}}{{{{\sin }^2}\theta }} - \frac{{{y^2}}}{{{{\cos }^2}\theta }} = 1 \cr
& {\text{or, }}{x^2}{\text{cose}}{{\text{c}}^2}\theta - {y^2}{\sec ^2}\theta = 1 \cr} $$