Equation of hyperbola with eccentricity $\frac32$

Question

A hyperbola whose transverse axis is along the major axis of the conic $\frac{2x^2}{3}+\frac{y^2}{4}=4$ and has vertices on the foci of this conic. If the eccentricity of the hyperbola is $\frac32$ then which of the following points does NOT lie on it? $(0,2)/(\sqrt5,2\sqrt2)/(\sqrt{10},2\sqrt3)/(5,2\sqrt3)$

Ellipse is $\frac{x^2}{6}+\frac{y^2}{16}=1$. Its eccentricity $\frac{\sqrt{10}}4$. Foci $(0,\pm\sqrt{10})$.

Let the equation of hyperbola be $-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Vertex$=b\frac32=\sqrt{10}\implies b=\frac{2\sqrt{10}}{3}$. And $a^2=b^2(e^2-1)=\frac{50}9$.

Thus, hyperbola is $-\frac{9x^2}{50}+\frac{9y^2}{40}=1$.

With this, no point is matching.

Answer 1

Equation of rotated hyperbola is$$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$$ Vertices of the hyperbola are on the foci of the ellipse, therefore $a=\sqrt{10}$

Eccentricity is $e=\frac{c}{b}=\frac32\to c=\frac32 b$

$b^2=c^2-a^2\to b^2=\frac{9}{4}b^2-10\to b^2=8$

Hyperbola equation is $$\frac{y^2}{10}-\frac{x^2}{8}=1$$

It is true that none of the points lie on the hyperbola. See the graph below.

The red hyperbola is what they had in mind (blue points are the given points and one of them does not lie on the red hyperbola), but its equation doesn't fit the given informations

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