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Questions  

Locus of the point of intersection of the normals to the parabola y2=16x which are at right angles is

a
y2=4(x−4)
b
y2=4(x−8)
c
y2=−4(x−12)
d
y2=−8(x−8)

detailed solution

Correct option is C

Let two perpendicular normals to the parabola y2=16xbe y=mx−8m−4m3  and  y=−1mx+8m+4m3Solving we get x=4m−1m2+12y=−4m−1mEliminating m we get the required locus asy2=4(x−12).

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