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. $(\sqrt{5},2\sqrt{2})$
• B. $(0,2)$
• C. $(5,2\sqrt{3})$
• D. $(\sqrt{10},2\sqrt{3})$

Answer 1

Answer: (C)

$(5,2\sqrt{3})$

$\frac{x^2}{12}+\frac{y^2}{16}=1$

$e=\sqrt{1-\frac{12}{16}}=\frac{1}{2}$

Foci$=(0.\pm be)$ $(0,2)$ and $(0,-2)$

So, transverse axis of hyperbola $=2b=4$

$\Rightarrow b=2$

and $a^2=1^2(e^2-1)$

$\Rightarrow a^2=4(\frac{9}{4}-1)$

$\Rightarrow a^2=5$

$\therefore$ It's equation is $\frac{x^2}{5}-\frac{y^2}{4}=-1$

$(5,2\sqrt{3})$ does not satisfy above equation.