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Question

Given slope m, find the equation of normal having slope m to the parabola y2=4ax


A

y=mx+2amam3

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B

y=mx2amam3

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C

y=mx2amm3

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D

y=mx2am+am3

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Solution

The correct option is B

y=mx2amam3


Given value of normal is m. we know at point (h,k) slope of tangent is 2ak. Therefore slope of normal would be k2a.

We have, k2=4ah and m=k2a

Substituting for c in y = mx + c at point (h,k)

k = m(h) + c

2a(m)=m(k24a)+c

2amm((2am)24a)=c

we get c=2amam3

Therefore equation of normal with slope m for the given parabola would be y=mx2amam3


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