Question

A hyperbola, having the transverse axis of length 2 sin θ, is confocal with the ellipse 3x² + 4y² = 12. Then its equation is
(a) x²cosec²θ – y²sec²θ = 1
(b) x²sec²θ – y²cosec²θ = 1
(c) x²sin²θ – y²cos²θ = 1
(d) x²cos²θ – y²sin²θ = 1

Answer 1

Answer:

• x²cosec²θ – y²sec²θ = 1

Step-by-step explanation:

$\mathtt{The \: length \: of \: transverse \: axis \: = 2sinθ = \: 2a}$

$\qquad \qquad \qquad \implies \mathtt{a = \frac{ \cancel2sinθ}{\cancel2} }\\ \\ \qquad \qquad \qquad \implies \frak{a = \sin \theta}</p><p>$

$\tt \: also \: \: for \: \: ellipse \: \: 3 {x}^{2} + {4y}^{2} = 12$

or,

$\tt\frac{ {x}^{2} }{4} + \frac{ {y}^{2} }{3} = 1 \\ \tt \: where \: {a}^{2} = 4 \: and \: {b}^{2} = 3 \\ \\ e = \sqrt{1 - \frac{ {b}^{2} }{ {a}^{2} } } = \sqrt{1 - \frac{3}{4} } = \frac{1}{2} \\ \\ \because \tt focus \: of \: ellipse \: = \bigg \lgroup2 \times \frac{1}{2} ,0 \bigg \rgroup \\ \\ \: \qquad \qquad \implies \: (1,0)$

As hyperbola is confocal with ellipse, focus of hyperbola = (1,0)

$\implies \: ae = 1 \\ \implies \sin \theta \times e = 1 \\ \implies \: e = \csc \theta \\ \\ \qquad \: \because \: {b}^{2} = {a}^{2} ( {e}^{2} - 1) \\ \\ \qquad = { \sin }^{2} \theta( { \csc }^{2} \theta - 1) = { \cos}^{2} \theta$

$\because \mathcal E \sf{quation \: of \: hyperbola \: is} \\ \\ \qquad \qquad \sf \: \frac{ {x}^{2} }{ { \sin }^{2} \theta } - \frac{ {y}^{2} }{ { \cos }^{2} \theta } = 1$

$\qquad \qquad \bigg \lgroup \sf cosec \theta = \frac{1}{ \sin \theta} \\ \qquad \qquad \: \qquad \sf sec \theta = \frac{1}{ \cos \theta } \bigg \rgroup$

$\implies \underline{\mathcal{\red{ {x}^{2} { \csc }^{2} \theta - {y}^{2} { \sec }^{2} \theta } = 1}}$

Answer 2

Answer:

(a) x²cosec²θ – y²sec²θ = 1