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If the normal at one end of the latus rectum of the parabola $y^2=16 x$ meets the $X$-axis at the point $P$, then the length of the chord passing through $P$ and perpendicular to the normal is
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The correct answer is:
$32 \sqrt{2}$
We have, $y^2=16 x$
Coordinate of latusrectum $(4,8)$
Equation of normal at $(4,8)$ is $y=-x+12$
It cuts $\mathrm{x}$-axis at $(12,0)$
Equation of chord passing through $(12,0)$ and perpendicular of normal is $y=x-12$
Put the value of $y$ in $y^2=16 x$, we get
$$
\begin{gathered}
(12-x)^2=16 x \\
144-24 x+x^2=16 x \\
x^2-40 x+144=0 \\
(x-36)(x-4)=0 \\
x=4,36 \\
\therefore \quad y=4-12=-8 \text { and } y=36-12=24 \\
\therefore \text { Length of chord }=\sqrt{(36-4)^2+(24+8)^2} \\
=\sqrt{32^2+32^2}=32 \sqrt{2}
\end{gathered}
$$
Coordinate of latusrectum $(4,8)$
Equation of normal at $(4,8)$ is $y=-x+12$
It cuts $\mathrm{x}$-axis at $(12,0)$
Equation of chord passing through $(12,0)$ and perpendicular of normal is $y=x-12$
Put the value of $y$ in $y^2=16 x$, we get
$$
\begin{gathered}
(12-x)^2=16 x \\
144-24 x+x^2=16 x \\
x^2-40 x+144=0 \\
(x-36)(x-4)=0 \\
x=4,36 \\
\therefore \quad y=4-12=-8 \text { and } y=36-12=24 \\
\therefore \text { Length of chord }=\sqrt{(36-4)^2+(24+8)^2} \\
=\sqrt{32^2+32^2}=32 \sqrt{2}
\end{gathered}
$$
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