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Let $P$ represent the point $(3,6)$ on the parabola $y^2=12 x$. For the parabola $y^2=12 x$, if $l_1$ is the length of the normal chord drawn at $P$ and $l_2$ is the length of the focal chord drawn through $P$, then $\frac{l_1}{l_2}=$
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The correct answer is:
$2 \sqrt{2}$
We know that, length of normal of parabola $y^2=4 a x$ at $\left(a t^2, 2 a t\right)=8 a \sqrt{t^2+1}$ and length of focal chord $=a\left(t+\frac{1}{t}\right)^2$.
Point $P(3,6)$ lie on parabola $y^2=12 x$
$\therefore$ Here $a=3, t=1$
Hence $l_1=8 \times 3 \sqrt{1+1}=24 \sqrt{2}$
$$
\begin{aligned}
& l_2=3(1+1)^2=12 \\
& \frac{l_1}{l_2}=\frac{24 \sqrt{2}}{12}=2 \sqrt{2}
\end{aligned}
$$
Point $P(3,6)$ lie on parabola $y^2=12 x$
$\therefore$ Here $a=3, t=1$
Hence $l_1=8 \times 3 \sqrt{1+1}=24 \sqrt{2}$
$$
\begin{aligned}
& l_2=3(1+1)^2=12 \\
& \frac{l_1}{l_2}=\frac{24 \sqrt{2}}{12}=2 \sqrt{2}
\end{aligned}
$$
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