Equation of a normal to the parabola y2 = 32x passing through its focus is

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x=0

y=0

x+y−8=0

x−y−8=0

answer is B.

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Detailed Solution

Equation of a normal to the parabola is y=mx−2am−am3,4a=32⇒a=8which passes through the focus (8, 0)If 0=8m−16m−8m3=0 ⇒m3+m=0⇒mm2+1=0⇒m=0and the required equation of the normal is y = 0.

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