Question

Let the hyperbola \(H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) pass through (\(2\sqrt2,-2\sqrt2\) ). A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?

• A. \(2\sqrt3,3\sqrt2\)
• B. \(3\sqrt3,-6\sqrt2\)
• C. \(\sqrt3,-\sqrt6\)
• D. \(3\sqrt6,6\sqrt2\)

Answer 1

The Correct Option is B

\(\begin{array}{l} H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \end{array}\)

Focus of parabola: (ae, 0)

Directrix: x = – ae.

Equation of parabola ≡y² = 4aex

Length of latus rectum of parabola = 4ae

\(\begin{array}{l}\text{Length of latus rectum of hyperbola} =\frac{2\cdot b^2}{a} \end{array}\)

as given,

\(\begin{array}{l} 4ae=\frac{2b^2}{a}\cdot e\end{array}\)

\(\begin{array}{l} 2=\frac{b^2}{a^2}\cdots \left(i\right)\end{array}\)

\(\begin{array}{l} \because \text{H passes through} \left(2\sqrt{2},-2\sqrt{2}\right)\Rightarrow \frac{8}{a^2}-\frac{8}{b^2}=1\cdots \left(ii\right)\end{array}\)

From (i) and (ii) a² = 4 and b² = 8

\(\begin{array}{l}\Rightarrow e=\sqrt{3}\end{array}\)

\(\begin{array}{l} \Rightarrow\text{Equation of parabola is}~ y^2=8\sqrt{3}x\end{array}\)