# Finding the equation of the normal to the parabola $y^2=4x$ that passes through $(9,6)$

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) $$\;y − x + 3 = 0$$

(B) $$\;y + 3x − 33 = 0$$

(C) $$\;y + x − 15 = 0$$

(D) $$\;y − 2x + 12 = 0$$

My attempt: Let $$(h,k)$$ be the point on parabola where normal is to be found out. Taking derivative, I get the slope of the normal to be $$\frac{-k}{2}$$. Since the normal passes through $$(9,6)$$, so, the equation of the normal becomes:$$y-6=\frac{-k}{2}(x-9)$$$$\implies \frac{kx}{2}+y=\frac{9k}{2}+6$$

By putting $$k$$ as $$2,-2,-4$$ and $$6$$, I get normals mentioned in $$A,B,C$$ and $$D$$ above (not in that order).

But the answer is given as $$A,B$$ and $$D$$. What am I doing wrong?

• Isn't slope of normal $\frac {-k}{2h}$ ? Jul 5 '20 at 9:35

In addition to having slope $$-k/2$$ the normal must also pass through the point of contact $$(h,k)$$. The line in option $$C$$ does not pass through the point of contact for $$k=2$$ which is $$(1,2)$$. Your equation is the equation of a line having the slope of a normal at point $$(h,k)$$ on the parabola and passing through $$(9,6)$$. It is not necessarily a normal because you didn't make it pass through the point of contact.

The equation of the normal is $$\color{red}{y=\frac{-kx}{2}+\frac{9k}{2}+6}$$. The point $$(h,k)$$ lies on it so we get $$k=\frac{-kh}{2}+\frac{9k}{2}+6.$$ But $$k^2=4h$$ (since $$(h,k)$$ lies on the parabola as well), so we get $$k^3-28k-48=0 \implies (k-6)(k+2)(k+4)=0 \implies k=6,-2,-4.$$ These value can be used to find the equation of the normal as: \begin{align*} y&=-3x+33\\ y &=x-3\\ y &=2x-12 \end{align*}

Use parametric equation : any point $$P(t^2,2t)$$ on the given parabola

The gradient of the tangent $$=\dfrac4{2y}_{\text{(t^2,2t)}}=\dfrac1t$$

So, the gradient of the normal $$=-t$$

So, the equation of the normal $$\dfrac{y-2t}{x-t^2}=-t \implies xt+y=2t+t^3$$

Here $$2t=6$$

The general equation for a normal to the parabola $$y^2=4ax$$ is

$$y=mx-2am-am^3$$

Putting $$a=1$$ and passing it through $$(9,6)$$, we have $$6=9m-2m-m^3$$ Solving the above cubic for $$m$$ yields three values

$$m=1,2 \text{ or } -3$$

Now substitute that back, and voilà!