First slide

Hyperbola in conic sections

Question

If the vertex of a hyperbola bisects the distance between its center and the corresponding focus, then the ratio of the square of its conjugate axis to the square of its transverse axis is

Moderate

A

2
B

4
C

6
D

3

Solution

Let the hyperbola be $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$Then, 2 a = ae, i.e., e = 2 . Therefore,$
$\begin{aligned} \frac{b^{2}}{a^{2}} = e^{2} - 1 = 3 \\ or \frac{(2 b)^{2}}{(2 a)^{2}} = 3 \end{aligned}$

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