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

Answer: (B)

$3\sqrt3,-6\sqrt2$

Answer 2

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

Focus of parabola: (ae , 0)

Directrix: x = – ae.

Equation of parabola ≡ y 2 = 4 aex

Length of latus rectum of parabola = 4 ae

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

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) a 2 = 4 and b 2 = 8

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

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