Question

A hyperbola whose transverse axis is along the major axis of the conic, $\frac{x^2}{3}+\frac{y^2}{4}=4$ and has vertices at the foci of this conic. If the eccentricity of the hyperbola is $\frac{3}{2}$, then which of the following points does NOT lie on it?

• A. $(5,2\sqrt{3})$
• B. $(0,2)$
• C. $(\sqrt{5},2\sqrt{2})$
• D. $(\sqrt{1}0,2\sqrt{3})$

Answer 1

The correct option is A $(5,2\sqrt{3})$

For the ellipse $\frac{x^2}{12}+\frac{y^2}{16}=1$,
$a^2=12,b^2=16$
$e^2=1-\frac{a^2}{b^2}(\because b>a)$
$e=\frac{1}{2}$
So, the foci are $(0,\pm 2)$
So, the vertices of the hyperbola are $(0,\pm 2)$


Let the equation of hyperbola be $\frac{y^2}{p^2}-\frac{x^2}{q^2}=1$
$\Rightarrow p=2$
Given that $e=\frac{3}{2}$
$\Rightarrow q=\sqrt{5}$

$\Rightarrow \frac{y^2}{4}-\frac{x^2}{5}=1$
The point $(5,2\sqrt{3})$ does not lie on the hyperbola.