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If $\overrightarrow{A B}$ is the focal chord of the parabola $y^2=16 x$ and $A=(1,-4)$, then the equation of the normal to the parabola at the point $\mathrm{B}$ is
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$2 x+y-48=0$
If parameter of $A$ and $B$ are $t_1, t_2$ then $t_1 \cdot t_2=-1$
$\begin{aligned} & \Rightarrow a t_1{ }^2=1 \Rightarrow 2 a t_1=-4 \Rightarrow t_1=-\frac{1}{2} \text { and } a=4 \\ & \therefore \quad t_2=2 \\ & B\left(a t_2^2, 2 a t_2\right) \equiv(16,16)\end{aligned}$
Equation of normal : $y+t_2 x=2 a t_2+a t_2^3$
$\Rightarrow y+2 x=16+32 \Rightarrow 2 x+y-48=0$
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