Animated Solution for Mathematics - Conic Sections: 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?

Unveiling the Hyperbola

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a journey to understand the elegant, sweeping curves of a hyperbola.

Imagine you are standing on a vast, two-dimensional plane. You have two fixed points, the vertices, and a hidden, gravitational anchor, the focus. Our goal is to map the path of a point that dances around these constraints.

Look closely at the information provided. We are given the vertices at $(-2,0)$ and $(2,0)$.

In the world of conic sections, the vertices are the gateways of the hyperbola. They are the points where the curve is closest to the center. Because these points are symmetric about the origin, we immediately know that the center of our hyperbola is at $(0,0)$.

This is a standard horizontal hyperbola, which means its equation will take the form:

By comparing our given vertices $(\pm 2,0)$ to the standard form $(\pm a,0)$, we can see with absolute clarity that $a=2$. Thus, $a^2=4$. We have our first anchor.

Now, let us turn our attention to the focus. The problem tells us that one focus is at $(-3,0)$.

The focus is the 'gravity' of the hyperbola, the point that defines its stretch, or eccentricity, $e$. The standard coordinates for the foci are $(\pm ae,0)$.

Since our focus is at $(-3,0)$, we know that $ae=3$. We already know that $a=2$. Therefore, $2e=3$, which gives us an eccentricity of:

Notice that $e>1$, which is the hallmark of a hyperbola. It is a beautiful confirmation that our geometry is sound.

We are almost there. To complete our equation, we need the value of $b^2$. There is a fundamental relationship that binds $a$, $b$, and $e$ together in the hyperbola's soul:

Let us substitute our known values into this elegant expression. We have $a^2=4$ and $e^2=(\frac{3}{2}{)}^2=\frac{9}{4}$.

So, $b^2=4(\frac{9}{4}-1)$. Simplifying the bracket, we get $\frac{9}{4}-\frac{4}{4}=\frac{5}{4}$. Multiplying this by $4$, the fours cancel out, leaving us with $b^2=5$.

The equation of our hyperbola is now fully revealed:

Now, we must test the options. This is where we verify our work. We are looking for the point that does not lie on the curve.

For the first point, $(4,\sqrt{15})$, we substitute $x=4$ and $y=\sqrt{15}$ into our equation:

It works! This point is on the hyperbola.

For the second point, $(-6,2\sqrt{10})$, we substitute $x=-6$ and $y=2\sqrt{10}$:

It works again!

Now, the third point, $(6,5\sqrt{2})$. Let us substitute $x=6$ and $y=5\sqrt{2}$:

Wait! This is not $1$. This point does not satisfy the equation. It lies on the conjugate hyperbola, not the one we defined.

Finally, the fourth point, $(2\sqrt{6},5)$. Substituting $x=2\sqrt{6}$ and $y=5$:

This point also lies on the hyperbola.

We have successfully navigated the geometry, derived the equation, and verified the points. The point $(6,5\sqrt{2})$ is the outlier, the one that does not belong to our curve.

Remember, in JEE Advanced, it is not just about finding the answer; it is about understanding the path. You have mastered the hyperbola today. Keep that curiosity burning!