If the vertices of a hyperbola be at $$(-2, 0)$$ and $$(2, 0)$$ and one of its foci be at $$(-3, 0)$$, then which one of the following points does not lie on this hyperbola?

We are told that the vertices of the required hyperbola are at $$(-2,0)$$ and $$(2,0)$$. For any hyperbola with its transverse (real) axis along the $$x$$-axis, the standard form is

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,$$

where $$\left(\pm a,0\right)$$ are the vertices. Comparing this fact with the given vertices, we at once see that

$$a=2,\qquad a^{2}=4.$$

Next, one focus of the hyperbola is stated to be at $$(-3,0)$$. In the same standard orientation, the foci are at $$\left(\pm c,0\right)$$, so we identify

$$c=3,\qquad c^{2}=9.$$

A basic hyperbola identity connects the semi-transverse axis $$a$$, the semi-conjugate axis $$b$$ and the focal distance $$c$$:

$$c^{2}=a^{2}+b^{2}\qquad\text{(for a hyperbola).}$$

Substituting the already found values, we get

$$9 = 4 + b^{2},$$

so

$$b^{2}=9-4=5.$$

Hence, the complete Cartesian equation of the hyperbola is

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

Now we shall test each of the four given points one by one by substituting their coordinates into this equation. A point lies on the curve only when the left-hand side equals $$1$$.

Option A: $$(6,\,5\sqrt{2})$$

We have

$$\frac{x^{2}}{4}=\frac{6^{2}}{4}=\frac{36}{4}=9,$$

and

$$\frac{y^{2}}{5}=\frac{(5\sqrt{2})^{2}}{5}=\frac{25\cdot 2}{5}=\frac{50}{5}=10.$$

So the left-hand side becomes

$$9-10=-1\neq 1.$$

Therefore $$(6,\,5\sqrt{2})$$ does not satisfy the equation and is not on the hyperbola.

Option B: $$(-6,\,2\sqrt{10})$$

Compute

$$\frac{x^{2}}{4}=\frac{(-6)^{2}}{4}=\frac{36}{4}=9,\qquad \frac{y^{2}}{5}=\frac{(2\sqrt{10})^{2}}{5}=\frac{4\cdot 10}{5}=8.$$

The result is

$$9-8=1,$$

so this point indeed lies on the hyperbola.

Option C: $$(2\sqrt{6},\,5)$$

Compute

$$\frac{x^{2}}{4}=\frac{(2\sqrt{6})^{2}}{4}=\frac{4\cdot 6}{4}=6,\qquad \frac{y^{2}}{5}=\frac{5^{2}}{5}=\frac{25}{5}=5.$$

Thus

$$6-5=1,$$

so this point also lies on the curve.

Option D: $$(4,\,\sqrt{15})$$

Compute

$$\frac{x^{2}}{4}=\frac{4^{2}}{4}=\frac{16}{4}=4,\qquad \frac{y^{2}}{5}=\frac{(\sqrt{15})^{2}}{5}=\frac{15}{5}=3.$$

Hence

$$4-3=1,$$

so this point, too, satisfies the hyperbola’s equation.

Among the four points, only Option A fails to satisfy $$\dfrac{x^{2}}{4}-\dfrac{y^{2}}{5}=1$$ and hence does not belong to the hyperbola.

Hence, the correct answer is Option A.