Let E be the ellipse x²/16 + y²/9 = 1. For any three distinct points P, Q and Q' on E, let M(P, Q) b

Q: Wait, why does the point $P$ disappear from the final expression? Did I do something wrong?

You are doing perfectly! This is actually the most elegant part of the problem. When you calculate the distance between the midpoints, you are subtracting $\frac{\vec{p} + \vec{q}}{2}$ and $\frac{\vec{p} + \vec{q}'}{2}$. The $\vec{p}$ terms cancel out exactly, leaving you with $\frac{1}{2}(\vec{q} - \vec{q}')$. This tells us that the distance between the midpoints is independent of the position of $P$. It only depends on the distance between $Q$ and $Q'$.

Q: Why can't we just use energy conservation or some other physics principle here?

That is a great instinct for a physics student, but this is a pure geometry problem! In JEE Advanced, you must choose the right tool for the job. Here, the tool is coordinate geometry and vector algebra. Energy conservation applies to systems changing over time, but here, the points are static geometric entities. Stick to the properties of the ellipse and the Midpoint Theorem.

Q: Is the major axis always the longest chord in an ellipse?

Yes, absolutely. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$), the major axis is the longest diameter, and its length is $2a$. Any other chord will be shorter than this. Since we want to maximize the distance between $Q$ and $Q'$, we must choose them to be the endpoints of the major axis.

Question

Let $E$ be the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$. For any three distinct points $P, Q$ and $Q'$ on $E$, let $M(P, Q)$ be the mid-point of the line segment joining $P$ and $Q$, and $M(P, Q')$ be the mid-point of the line segment joining $P$ and $Q'$. Then the maximum possible value of the distance between $M(P, Q)$ and $M(P, Q')$, as $P, Q$ and $Q'$ vary on $E$, is ___.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a coordinate plane, and on it, a perfect, smooth ellipse is drawn, defined by the equation:

$$ \frac{x^2}{16} + \frac{y^2}{9} = 1 $$

This is not just a shape; it is a constraint. We have three points, $P$, $Q$, and $Q'$, dancing along this boundary. The problem asks us to find the maximum distance between the midpoints of the segments $PQ$ and $PQ'$.

At first glance, this might seem like a nightmare of algebraic complexity. You might be tempted to write out the coordinates of $P$, $Q$, and $Q'$ using parametric forms like $(4\cos	heta, 3\sin	heta)$ and start grinding through the distance formula. But stop; in the world of JEE Advanced, the most beautiful solutions are often the ones that bypass the brute force.

### The Power of the Midpoint Theorem

Let us shift our perspective. Instead of coordinates, let us look at the triangle $	riangle PQQ'$. We have two sides, $PQ$ and $PQ'$, sharing a common vertex $P$.

Geometry offers us a gift here: the **Midpoint Theorem**. This theorem states that the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is exactly half its length.

In our case, the third side is the segment $QQ'$. Therefore, the distance between our two midpoints must be exactly:

$$ \frac{1}{2} |QQ'| $$

### The Vectorial Shortcut

If you are still skeptical, let us verify this with the cold, hard logic of vectors. Let the position vectors of our points be $\vec{p}$, $\vec{q}$, and $\vec{q}'$.

The midpoint of $PQ$ is $\vec{M_1} = \frac{\vec{p} + \vec{q}}{2}$, and the midpoint of $PQ'$ is $\vec{M_2} = \frac{\vec{p} + \vec{q}'}{2}$. The distance between these midpoints is the magnitude of their difference:

$$ |\vec{M_1} - \vec{M_2}| = \left| \frac{\vec{p} + \vec{q}}{2} - \frac{\vec{p} + \vec{q}'}{2} ight| $$

Look closely at the algebra. The $\vec{p}$ terms subtract to zero! We are left with:

$$ \left| \frac{\vec{q} - \vec{q}'}{2} ight| = \frac{1}{2} |\vec{q} - \vec{q}'| $$

This confirms our geometric intuition: the distance between the midpoints is simply half the distance between $Q$ and $Q'$. The position of $P$ is completely irrelevant.

### Maximizing the Chord

Now, the problem transforms. We no longer care about $P$. We only care about maximizing the distance $|QQ'|$, where $Q$ and $Q'$ are any two points on the ellipse.

What is the longest possible distance between any two points on an ellipse? It is the length of the major axis. For our ellipse:

$$ \frac{x^2}{16} + \frac{y^2}{9} = 1 $$

We compare this to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We see that $a^2 = 16$, which means the semi-major axis $a = 4$. The length of the major axis is:

$$ 2a = 2(4) = 8 $$

Thus, the maximum distance between $Q$ and $Q'$ is $8$.

### The Final Victory

We have reached the finish line. We know the maximum distance between the midpoints is $\frac{1}{2}$ of the maximum distance between $Q$ and $Q'$.

Substituting our value, we get:

$$ \frac{1}{2} 	imes 8 = 4 $$

It is a moment of pure mathematical satisfaction. We started with a complex-looking problem involving three moving points and arrived at a simple, elegant constant. The **maximum distance is 4**.

Enter Numerical Value:

Visualized Solution

Visualizing the Ellipse and Points

Locating the Midpoints

The Midpoint Theorem

Vector Setup for Distance

Simplifying the Distance

Condition for Maximum Distance

Maximum Distance on an Ellipse

Finding the Semi-Major Axis

Calculating Major Axis Length

Final Midpoint Distance

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Power of the Midpoint Theorem

The Vectorial Shortcut

Maximizing the Chord

The Final Victory

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