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Question

Find the equation of tangents to the curve y=x3+2x4, which are perpendicular to line x+14y+3=0.

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Solution

Assuming the point of contact be (x1,y1).

As it (x1,y1) lies on y=x3+2x4.

y1=x31+2x14 ............(1)

Given curve,
y=x3+2x4

Differentiating the curve w.r.t. x, we get

dydx=3x2+2

dydx](x1,y1)=3x21+2 [ slope of the tangent at(x1,y1)]

Since the tangent at (x1,y1) is perependicular to the line x+14y+3=0

Therefore, slope of tangnet at x1,y1× slope of line = -1

dydx](x1,y1)×(114)=1

(3x21+2)(114)=1

x1=±2

Now, when x1=2,y1=(2)3+2(2)4=8

When x1=2,y1=(2)3+2(2)4=16

So, the coordinates of the point of contact are (2,8) and (-2,-16).

Equation of tangent for the coordinates (2,8) is:

y8=14(x2)

14xy20=0

Equation of tangent for the coordinates (-2,-16) is:

y+16=14(x+2)

14xy+12=0.

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