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Question

Find the equation of the tangent line to the curve y=x22x+7 which is parallel to the line 5y-15x=13.

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Solution

The equation of the given curve is y=x22x+7

On differentiating w.r.t x, we get dydx=2x2

(b) The equation of the given line is 5y-15x=13 y=3x+135

This is of the form y=mx+c

Slope of the line is 3.

If a tangent is perpendicular to the line 5y-15x=13, Then the slope of the tangent =13

[product of slopes, m1× m2=1m2=1m1]

2x2=132x=213=53x=56

When x=56, then from Eq (i), we get

y=(56)22(56)+7=253657+7=2560+25236=21736

The point on the given curve at which tangent is parallel to given line is (56,21736) and the equation of the tangent is

y21736=13(x56)y21736=x3+51812x+36y227=0

Hence, the equation of the tangent line to the given curve which is perpendicular to the line 5y-15x=13 is 36y+12x-227=0.


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