If overrightarrow{AB} is the focal chord of t | vedclass

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(B) The equation of the parabola is $y^2=16x$,which is of the form $y^2=4ax$. Thus,$4a=16$,so $a=4$.Let the parameters of points $A$ and $B$ be $t_1$

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$2x+y-48=0$

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(B) The equation of the parabola is $y^2=16x$,which is of the form $y^2=4ax$. Thus,$4a=16$,so $a=4$.Let the parameters of points $A$ and $B$ be $t_1$ and $t_2$ respectively. Since $AB$ is a focal chord,$t_1 \cdot t_2 = -1$.For point $A(1, -4)$,we have $at_1^2 = 1$ and $2at_1 = -4$.Substituting $a=4$,we get $4t_1^2 = 1$ $\Rightarrow t_1^2 = 1/4$ $\Rightarrow t_1 = -1/2$ (since $y Since $t_1 \cdot t_2 = -1$,we have $(-1/2) \cdot t_2 = -1$,which gives $t_2 = 2$.Point $B$ is $(at_2^2, 2at_2) = (4(2)^2, 2(4)(2)) = (16, 16)$.The equation of the normal to the parabola $y^2=4ax$ at point $t$ is $y + tx = 2at + at^3$.For $t_2 = 2$ and $a=4$,the equation of the normal at $B$ is $y + 2x = 2(4)(2) + 4(2)^3$.$y + 2x = 16 + 32$.$y + 2x = 48$.Therefore,the equation of the normal is $2x + y - 48 = 0$.

If $\overrightarrow{AB}$ is the focal chord of the parabola $y^2=16x$ and $A=(1,-4)$,then the equation of the normal to the parabola at the point $B$ is

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