Animated Solution for Mathematics - Conic Sections: Let $O$ be the vertex and $Q$ be any point on the parabola, $x^2=8y$. If the point $P$ divides the line segment $OQ$ internally in the ratio $1:3$, then locus of $P$ is:

Imagine you are standing at the origin of a coordinate plane, looking up at a beautiful, symmetric curve: the parabola $x^2=8y$. This is not just a static shape; it is a path.

A point $Q$ is sliding along this curve, tracing its trajectory. We are interested in a point $P$ that is tethered to this moving point $Q$, always maintaining a specific relationship with the origin $O$.

This is the essence of a locus problem—finding the hidden path of a point that is constrained by the movement of another.

The Parametric Key

To solve this, we must first tame the movement of $Q$. The equation $x^2=8y$ is in the standard form $x^2=4ay$, where $4a=8$, giving us $a=2$.

Instead of dealing with $x$ and $y$ as independent variables, we introduce a parameter $t$. By setting $x=4t$, we find that $y$ is:

Thus, any point $Q$ on our parabola can be elegantly represented as $Q(4t,2t^2)$. As $t$ varies, $Q$ sweeps across the entire parabola, serving as our bridge between geometry and algebra.

The Section Formula Bridge

Now, consider the line segment $OQ$. $O$ is the origin $(0,0)$, and $Q$ is our moving point $(4t,2t^2)$.

We are looking for a point $P(h,k)$ that divides this segment in the ratio $1:3$. The section formula is our best friend here, as it provides the coordinates of $P$ via the weighted average of the endpoints.

For our ratio $1:3$, the coordinates of $P$ are:

The terms involving $0$ vanish, leaving us with simple expressions for $h$ and $k$ in terms of $t$.

The Final Elimination

We have $h=t$ and $k=\frac{t^2}{2}$. We want the locus of $P$, which means we need a direct relationship between $h$ and $k$ that does not depend on the parameter $t$.

Since $h=t$, we can simply substitute $h$ for $t$ in the second equation:

Rearranging this, we get $h^2=2k$. Replacing $h$ and $k$ with the standard variables $x$ and $y$, we arrive at the final locus: $x^2=2y$.

As you advance in your JEE preparation, you will start to see patterns. For any parabola $x^2=4ay$, if a point $P$ divides the segment from the vertex to a point on the curve in the ratio $1:n$, the locus is always another parabola:

In our case, $4a=8$ and $n=3$. Therefore, the locus is:

This shortcut is a powerful tool to verify your work in seconds. Keep exploring, keep visualizing, and remember: every complex problem is just a series of simple, elegant steps waiting to be discovered.