Animated Solution for Mathematics - Conic Sections: The locus of the mid-point of the line segment joining the focus of the parabola $y^2=4ax$ to a moving point of the parabola, is another parabola whose directrix is:

A Geometric Journey

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are tracing a path.

We are looking at the parabola $y^2=4ax$ and asking a profound question: if we take the focus $S(a,0)$ and connect it to a point $P$ that dances along the curve, what path does the midpoint $M$ of that segment trace? This is the essence of a locus problem—it is about capturing the ghost of a movement in a static equation.

The Setup

Imagine you are standing at the focus $S(a,0)$ of the parabola $y^2=4ax$. This point is your anchor.

Now, imagine a point $P$ sliding along the curve. To describe $P$ effectively, we use the magic of parametric coordinates: $P(a{t}^2,2at)$.

Because in the world of JEE, parametric coordinates are your best friend. They transform the square-root-heavy Cartesian equation into a clean, polynomial-based playground.

As $t$ varies, $P$ traces the entire parabola. Our midpoint $M(h,k)$ is simply the average of $S$ and $P$. Mathematically, this is the midpoint formula:

The Parametric Bridge

Now, we have two equations: $h=\frac{a(t^2+1)}{2}$ and $k=at$. Our goal is to find the relationship between $h$ and $k$.

This is the heart of the problem: we must eliminate the parameter $t$. Look at the equation for $k$. It is beautifully simple: $k=at$.

This implies $t=\frac{k}{a}$. This is our bridge. By isolating $t$, we have found a way to express the movement of $P$ entirely in terms of the coordinates of $M$.

The Transformation

Now, we substitute $t=\frac{k}{a}$ into our equation for $h$. This is where the algebra gets exciting.

We have:

Let's expand this carefully:

Multiplying by $2$, we get $2h=\frac{k^2}{a}+a$. Subtracting $a$ from both sides, we arrive at $2h-a=\frac{k^2}{a}$. Finally, multiplying by $a$, we get:

The Revelation

Let's rewrite this as:

If we replace $(h,k)$ with $(x,y)$ to represent the general locus, we get:

This is clearly a parabola! It is a shifted version of our original curve. The vertex has moved from $(0,0)$ to $(\frac{a}{2},0)$.

The Directrix

Finally, we find the directrix. For a standard parabola $Y^2=4AX$, the directrix is $X=-A$.

Here, $X=x-\frac{a}{2}$ and $A=\frac{a}{2}$. So, $x-\frac{a}{2}=-\frac{a}{2}$.

The terms cancel out, leaving us with $x=0$. The directrix is the $y$-axis!

Isn't that elegant? A simple midpoint operation on a parabola results in another parabola, and its directrix lands perfectly on the $y$-axis. Keep practicing, keep visualizing, and remember: every equation tells a story.