Animated Solution for Mathematics - Conic Sections: Let $L$ be a normal to the parabola $y^2=4x$. If $L$ passes through the point $(9,6)$, then $L$ is given by

A Journey into Parabolas

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant architecture of the parabola.

When we talk about a normal to a parabola, we are talking about a line that is perpendicular to the tangent at a specific point. It is a line that cuts through the curve with absolute precision.

Our mission is to find the equations of the normals to the parabola $y^2=4x$ that pass through the point $(9,6)$.

The Setup

First, let us ground ourselves. We are given the parabola $y^2=4x$.

By comparing this to the standard form $y^2=4ax$, we immediately identify that $4a=4$, which means our parameter $a=1$.

This parameter is the heartbeat of our parabola; it dictates its width and curvature. We are looking for lines that are normal to this curve and pass through the point $(9,6)$.

The Weaponry

To solve this, we need the right tool. We do not want to deal with the messy coordinates of the foot of the normal if we can avoid it.

Instead, we use the slope-form equation of a normal to a parabola. For the parabola $y^2=4ax$, the equation of a normal with slope $m$ is given by:

Since we know $a=1$, our equation simplifies beautifully to:

This equation is our gateway. It tells us that for any slope $m$, there is a corresponding normal line. But which of these lines pass through $(9,6)$?

The Intersection

This is where the magic happens. We force the line to pass through $(9,6)$ by substituting $x=9$ and $y=6$ into our equation.

This transforms our geometric problem into a purely algebraic one:

Let us simplify this. Combining the terms on the right, we get $6=7m-m^3$. Rearranging this into a standard cubic form, we arrive at:

This cubic equation is the key to the kingdom. It tells us that there are three possible slopes for the normals passing through our point. If we find the roots of this equation, we find our lines.

The Hunt for Roots

Solving a cubic can be daunting, but in JEE Advanced, there is almost always a hidden symmetry. Let us test simple integers.

If we try $m=1$, we get $1-7+6=0$. It works! So, $(m-1)$ is a factor.

Dividing the cubic by $(m-1)$, we get the quadratic $m^2+m-6=0$. Factoring this, we find $(m-2)(m+3)=0$.

Thus, our three slopes are $m=1$, $m=2$, and $m=-3$.

The Conclusion

Now, we simply plug these slopes back into our normal equation $y=mx-2m-m^3$:

1. For $m=1$: $y=1(x)-2(1)-(1{)}^3⟹y=x-3⟹\mathbf{y}-\mathbf{x}+3=0$.

2. For $m=2$: $y=2(x)-2(2)-(2{)}^3⟹y=2x-12⟹\mathbf{y}-2x+12=0$.

3. For $m=-3$: $y=-3(x)-2(-3)-(-3{)}^3⟹y=-3x+33⟹\mathbf{y}+3x-33=0$.

We have found all three! Notice the elegance: the sum of the slopes $1+2-3=0$.

This is not a coincidence; it is a fundamental property of co-normal points. You have mastered the geometry, the algebra, and the theory. Well done.