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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?

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

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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*}

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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 $2t=6$

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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!

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