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Question

Find the equation of the tangent to the curve y=3x2 which is parallel to the line 4x2y+5=0.
Further find the equation of the normal to the curve.

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Solution

Equation of tangent is dydx=123x2×3 at (h,k)slope of tangent=323h2 Given tangent is parallel to line 4x2y+5 so slope of tangent=slope of line

323h2=2

9=16(3h2)

3h=2+916

h=4148

(h1k) lies on curve so

k=3h2=3×41482=916=34

(h1k)=(4148,34)

eqn of tangent is y34=2(x4148)

4y3=848(48x41)

24y18=40x41

48x24y=23

eqn of normal is y34=12(x4148)

4y3=248(48x41)

96y72=48x+41

48x+96y=113


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