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 the conic. If the eccentricity of the hyperbola is $$\frac{3}{2}$$, then which of the following points does not lie on the hyperbola?

We are given the ellipse

$$\frac{x^{2}}{3}\;+\;\frac{y^{2}}{4}\;=\;4.$$

First we write it in the standard form $$\dfrac{x^{2}}{a^{2}}\;+\;\dfrac{y^{2}}{b^{2}}\;=\;1.$$ Dividing every term by $$4$$ we obtain

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

So we can read

$$a^{2}=12,\qquad b^{2}=16.$$

Because $$b^{2}>a^{2}$$ the major axis is the $$y$$-axis. For an ellipse the focal distance satisfies the formula

$$c^{2}=b^{2}-a^{2}.$$

Substituting $$b^{2}=16$$ and $$a^{2}=12$$ gives

$$c^{2}=16-12=4\;\;\Longrightarrow\;\;c=2.$$

Hence the foci of the ellipse are $$\bigl(0,\pm2\bigr).$$

According to the question, the hyperbola has its transverse axis along the same major axis (the $$y$$-axis) and its vertices are exactly these two focal points. Therefore the centre of the hyperbola is still the origin and the distance from the centre to each vertex is

$$a'=2.$$\

(Here and henceforth we use primes to denote the hyperbola’s parameters.)

The eccentricity of the hyperbola is given as

$$e'=\frac{3}{2}.$$

For any hyperbola the relation between eccentricity, semi-transverse axis and focal distance is

$$e'=\frac{c'}{a'}\quad\Longrightarrow\quad c'=e'\,a'.$$

Substituting $$e'=\dfrac{3}{2}$$ and $$a'=2$$ we get

$$c'=\frac{3}{2}\times2=3.$$

For a hyperbola the semi-conjugate axis $$b'$$ satisfies

$$c'^{2}=a'^{2}+b'^{2}.$$

Putting $$c'=3,\;a'=2$$ gives

$$9=4+b'^{2}\;\;\Longrightarrow\;\;b'^{2}=5.$$

Since the transverse axis is vertical, the standard equation of the required hyperbola is

$$\frac{y^{2}}{a'^{2}}-\frac{x^{2}}{b'^{2}}=1 \;\;\Longrightarrow\;\; \frac{y^{2}}{4}-\frac{x^{2}}{5}=1.$$

We now test each of the four given points in the equation

$$\frac{y^{2}}{4}-\frac{x^{2}}{5}=1.$$

Option A: $$\bigl(\sqrt{5},\,2\sqrt{2}\bigr)$$

$$ \frac{(2\sqrt{2})^{2}}{4}-\frac{(\sqrt{5})^{2}}{5} =\frac{8}{4}-\frac{5}{5}=2-1=1. $$ The equation is satisfied, so the point lies on the hyperbola.

Option B: $$\bigl(0,\,2\bigr)$$

$$ \frac{(2)^{2}}{4}-\frac{0^{2}}{5} =\frac{4}{4}-0=1-0=1. $$ The equation is satisfied, so this point also lies on the hyperbola.

Option C: $$\bigl(5,\,2\sqrt{3}\bigr)$$

$$ \frac{(2\sqrt{3})^{2}}{4}-\frac{(5)^{2}}{5} =\frac{12}{4}-\frac{25}{5}=3-5=-2\neq1. $$ The left-hand side is $$-2,$$ not $$1,$$ so this point does not lie on the hyperbola.

Option D: $$\bigl(\sqrt{10},\,2\sqrt{3}\bigr)$$

$$ \frac{(2\sqrt{3})^{2}}{4}-\frac{(\sqrt{10})^{2}}{5} =\frac{12}{4}-\frac{10}{5}=3-2=1. $$ Again the equation is satisfied, so the point is on the hyperbola.

Among the four choices, only Option C fails to satisfy the hyperbola’s equation.

Hence, the correct answer is Option C.