JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Hyperbola

The ellipse $3x^2+4y^2=12$ is $\frac{x^2}{4}+\frac{y^2}{3}=1$, with $c=\sqrt{4-3}=1$ and foci $(\pm 1,0)$. The hyperbola shares these foci and has transverse axis $\sqrt{2}$, so $a=\frac{1}{\sqrt{2}}$, $a^2=\frac{1}{2}$ and $b^2=c^2-a^2=\frac{1}{2}$, giving $x^2-y^2=\frac{1}{2}$. Testing $\left(\sqrt{\frac{3}{2}},\frac{1}{\sqrt{2}}\right)$ yields $\frac{3}{2}-\frac{1}{2}=1\ne \frac{1}{2}$, so the hyperbola does not pass through this point, whereas the others do satisfy the equation.