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Question

Find the equation of tangent to the curve y=3x2 which is parallel to the line 4x2y+=0. further find the equation of the normal to the curve.

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Solution

Consider the given equation of the curve,

y=3x2 ……(1)

differentiate with respect to x ,we get

dydx=323x2

But Given the equation,

4x2y=0 …..(2)

Slope of equation (2) is

=(4)2=2

Given tangent of equation (1) is parallel to equation (2), so

$\begin{align}

323x2=2

3x2=34

3x2=(34)2

3x=916

x=316

Value of x put in equation (2), we get.

4×316=2y

y=38

Now ,

y38=2(x316)

16x16y=3

This is required answer

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