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If the line $2 x+3 y+n=0$ is a tangent to the parabola $y^2=8 x$, then the equation of the normal drawn at the point $(2 n, 4 \sqrt{n})$ to the parabola $y^2=8 x$ is
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Verified Answer
The correct answer is:
$3 x+y-66=0$
Equation of tangent to $y^2=4 a x$ is $y=m x+\frac{a}{m}$
$$
\therefore y=m x+\frac{2}{m}
$$
Comparing with $2 x+3 y+n=0$
$$
\begin{aligned}
& m=\frac{-2}{3} \\
\Rightarrow & \frac{-n}{3}=\frac{2}{m} \Rightarrow n=\frac{-6}{m} \\
\therefore \quad & n=-6 \times \frac{3}{-2}=9 .
\end{aligned}
$$
$$
(2 n, 4 \sqrt{n})=(18,12)
$$
Equation of normal :
$$
\begin{aligned}
& y=-t x+2 a t+a t^3 \\
\because \quad & \left(a t^2, 2 a t\right)=(18,12) \\
& a=2 \Rightarrow t=3 \\
\therefore \quad & y=-3 x+12+54 \\
\Rightarrow & 3 x+y-66=0 .
\end{aligned}
$$
$$
\therefore y=m x+\frac{2}{m}
$$
Comparing with $2 x+3 y+n=0$
$$
\begin{aligned}
& m=\frac{-2}{3} \\
\Rightarrow & \frac{-n}{3}=\frac{2}{m} \Rightarrow n=\frac{-6}{m} \\
\therefore \quad & n=-6 \times \frac{3}{-2}=9 .
\end{aligned}
$$
$$
(2 n, 4 \sqrt{n})=(18,12)
$$
Equation of normal :
$$
\begin{aligned}
& y=-t x+2 a t+a t^3 \\
\because \quad & \left(a t^2, 2 a t\right)=(18,12) \\
& a=2 \Rightarrow t=3 \\
\therefore \quad & y=-3 x+12+54 \\
\Rightarrow & 3 x+y-66=0 .
\end{aligned}
$$
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