JEE Advanced · Mathematics · 14. Ellipse

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
$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$
B
$x^2 \sec ^2 \theta-y^2 \operatorname{cosec}^2 \theta=1$
C
$x^2 \sin ^2 \theta-y^2 \cos ^2 \theta=1$
D
$x^2 \cos ^2 \theta-y^2 \sin ^2 \theta=1$

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Answer & Solution

Correct Answer

(A)
$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$

Step-by-step Solution

Detailed explanation

The given ellipse is
\[
\begin{aligned}
& \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow \alpha=2, \beta=\sqrt{3} \\
& \therefore \quad 3=4\left(1-e^2\right) \Rightarrow e=\frac{1}{2} \\
&
\end{aligned}
\]
\[
\therefore \quad a e=1
\]
Hence the eccentricity $e_1$, of the hyperbola is given by
\[
\begin{array}{rlrl}
& & 1 & =e_1 \sin \theta \\
\Rightarrow & & e_1 & =\operatorname{cosec} \theta \\
\Rightarrow & b^2 & =\sin ^2 \theta\left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta
\end{array}
\]
Hence, the hyperbola is
\[
\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1 \text { or } x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1
\]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

If a tangent to a suitable conic (Column 1) is found to be

y = x + 8

and its point of contact is

(8, 16)

, then which of the following options is the only Correct combination?

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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

Answer & Solution

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