Animated Solution for Mathematics - Conic Sections: Find the point on the curve $4x^2+a^2{y}^2=4a^2,4<a^2<8$ that is farthest from the point $(0,-2)$.

Welcome, fellow traveler, to the beautiful world of coordinate geometry. Today, we are exploring the landscape of an ellipse defined by the equation $4x^2+a^2{y}^2=4a^2$.

To truly understand this shape, we must bring it into its standard form. By dividing the entire equation by $4a^2$, we reveal its true nature:

Here, we see an ellipse with a semi-major axis of $a$ and a semi-minor axis of $2$. Our mission is to find the point on this curve that is farthest from the point $Q(0,-2)$.

When dealing with ellipses, Cartesian coordinates $(x,y)$ can sometimes feel like a straightjacket. Instead, let us embrace the parametric form. Any point $P$ on this ellipse can be described by the angle $\theta$, where $P=(a\cos \theta ,2\sin \theta )$.

We want to maximize the distance $L$ between $P$ and $Q(0,-2)$. Calculating $L$ directly involves a square root, which is a recipe for algebraic headaches. Instead, we work with the square of the distance, $L^2$.

Since $L^2$ is a monotonic function of $L$, maximizing one maximizes the other. Our distance squared function becomes:

Expanding this, we get:

We are almost there, but we have two trigonometric functions, $\cos \theta$ and $\sin \theta$. Let us unify them using the identity ${\cos }^2\theta =1-{\sin }^2\theta$.

Let $s=\sin \theta$. Our function becomes $f(s)=a^2(1-s^2)+4(s+1{)}^2$. Expanding this, we arrive at a beautiful quadratic:

This is the heart of the problem. We are looking for the maximum of this parabola.

Here is where the JEE Advanced examiner tests your intuition. We have a parabola $f(s)=(4-a^2)s^2+8s+(a^2+4)$. Since $a^2>4$, the leading coefficient $(4-a^2)$ is negative, meaning the parabola opens downwards.

The vertex of this parabola occurs at:

Now, consider the constraint $4<a^2<8$. This implies $0<a^2-4<4$. If the denominator is less than $4$, then the fraction $\frac{4}{a^2-4}$ must be greater than $1$.

But wait! $s$ represents $\sin \theta$, and $\sin \theta$ is strictly confined to the interval $[-1,1]$. Our vertex lies outside the valid domain.

Because the vertex is to the right of our domain $[-1,1]$, and the parabola is opening downwards, the function $f(s)$ is strictly increasing as $s$ approaches $1$. Therefore, the maximum distance cannot be at the vertex; it must be at the boundary of our domain.

The maximum occurs at $s=1$. If $\sin \theta =1$, then $\theta =\frac{\pi }{2}$. Substituting this back into our parametric coordinates, we find:

We have found it. The point farthest from $(0,-2)$ is $(0,2)$. It is elegant, it is simple, and it is the result of careful, logical steps.