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 ?

Updated On: Aug 17, 2024
  • (0, 2)
  • $(\sqrt{5}$, 2 $\sqrt{2} )$
  • $(\sqrt{10}$, 2 $\sqrt{3})$
  • $( 5, 2 \sqrt{3} )$
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The Correct Option is D

Solution and Explanation



ellipse $\frac{x^{2}}{12}+\frac{y^{2}}{16}=1$
foci $(0, \pm$ be $)$
$e_{e}=\sqrt{1-\frac{12}{16}}=\frac{1}{2}$
for hyperbola


$e _{ H }=\frac{3}{2}$
equation $\Rightarrow \frac{ x ^{2}}{ a ^{2}}-\frac{ y ^{2}}{ b ^{2}}=-1$
$e _{ H }=\frac{3}{2}=\sqrt{1+\frac{ a ^{2}}{ b ^{2}}} $
$\Rightarrow \frac{9}{4}-1=\frac{ a ^{2}}{ b ^{2}}$
$\frac{ a ^{2}}{ b ^{2}}=\frac{5}{4} $
$\Rightarrow a ^{2}=5$
$\frac{ x ^{2}}{5}-\frac{ y ^{2}}{4}=-1$
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