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Question

Find the equations of tangent and normal to the parabola y2=4ax at the point (at2,2at).

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Solution

The equation of the given curve is y2=4ax.
Differentiating w.r.t. x, we get,

2ydydx=4a

dydx=2ay

Slope of tangent to the curve at (at2,2at) = 2a2at=1t

Therefore, equation of the tangent is

(y2at)=1t(xat2)

yt2at2=xat2

xty+at2=0

Slope of normal to the curve at (at2,2at) is t.

Therefore, equation of the normal is

(y2at)=t(xat2)

xt+y2atat2=0

xt+yat(2+t)=0

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