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Question

Find the equation of normal to the parabola y2=4ax at point (h,k) on the parabola


A

yk=ka(xh)

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B

yk=k2a(xh)

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C

yk=k2a(xh)

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D

yk=k2(xh)

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Solution

The correct option is B

yk=k2a(xh)


We know that the slope of the tangent of the given parabola at (h,k) is 2ak

. ðSlope of normal = 1(2ak)=k2a

y - k = k2a(xh)


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