Animated Solution for Mathematics - Conic Sections: Let $PQ$ be a chord of the parabola $y^2=12x$ and the midpoint of $PQ$ be at $(4,1)$. Then, which of the following point lies on the line passing through the points $P$ and $Q$?

Welcome, fellow traveler of the JEE journey. Today, we are not just solving a problem; we are uncovering the hidden elegance of coordinate geometry.

We are looking at a parabola, $y^2=12x$, and a chord $PQ$ that cuts through it. We are given the midpoint of this chord, $M(4,1)$, and we need to find the line that contains this chord.

Many students, upon seeing this, immediately reach for the slope-point form or try to find the coordinates of $P$ and $Q$ by solving a quadratic equation. While that is a valid path, it is the long, winding road. Today, I want to show you the highway.

In the world of conics, there is a beautiful, almost magical theorem known as the $T=S_1$ method. When you are given a conic section and a midpoint of a chord, you do not need to know the endpoints of the chord.

You only need the midpoint. The equation of the chord is given by the relation $T=S_1$.

Think of $T$ as the 'tangent-like' expression of the curve at the point $(x_1,y_1)$, and $S_1$ as the 'power' of the point, which is simply the value of the curve's equation when you plug in the coordinates of the midpoint. It is a transformation that turns a quadratic curve into a linear equation—a straight line.

Let us apply this to our parabola, $y^2-12x=0$. To construct $T$, we use the standard transformation rules for conics.

We replace $y^2$ with $y\cdot y_1$ and $2x$ with $(x+x_1)$. Since our equation is $y^2-12x=0$, we rewrite $12x$ as $6\cdot (2x)$.

Thus, our expression for $T$ becomes $y\cdot y_1-6(x+x_1)$. This is our linear operator.

Now, for $S_1$, we simply take the original equation $y^2-12x$ and substitute our midpoint $(4,1)$ into it. This gives us:

It is that simple. We are not doing complex calculus; we are performing a systematic substitution.

Now, let us execute the math. We have our midpoint $(x_1,y_1)=(4,1)$. Substituting these into our expressions, we get:

Expanding this, we get $y-6x-24$. On the other side, we calculate $S_1$:

Equating them, we have $y-6x-24=-47$. A quick rearrangement gives us $y-6x=-23$, or more elegantly:

This is the equation of the line passing through $P$ and $Q$. It is a straight line, perfectly defined by the midpoint we were given.

We are not done until we verify. The question asks which point lies on this line. We test the options.

Let us look at the point $(\frac{1}{2},-20)$. Substituting $x=\frac{1}{2}$ and $y=-20$ into our line equation $6x-y=23$, we get:

The left-hand side equals the right-hand side! The point lies perfectly on the line.

This is the beauty of the $T=S_1$ method. It is precise, it is elegant, and it is powerful. Keep this tool in your toolkit, and no chord problem will ever intimidate you again. You have mastered the geometry; now go forth and conquer the rest of your practice!