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(B) The given equation of the parabola is $y^{2}=16x$. Comparing this with the standard form $y^{2}=4ax$,we get $4a=16$,which implies $a=4$. It is a s

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intersect on the line $x+4=0$

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(B) The given equation of the parabola is $y^{2}=16x$. Comparing this with the standard form $y^{2}=4ax$,we get $4a=16$,which implies $a=4$. It is a standard property of parabolas that the tangents drawn at the extremities of any focal chord intersect at right angles on the directrix of the parabola. The equation of the directrix for the parabola $y^{2}=4ax$ is $x=-a$. Substituting $a=4$,the equation of the directrix is $x=-4$,which can be written as $x+4=0$. Therefore,the tangents intersect on the line $x+4=0$.

The tangents drawn at the extremities of a focal chord of the parabola $y^{2}=16x$:

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