Class 12
MATHS
Applications Of Derivatives

Find the equation of the tangent and the normal to the curve y = x^(2) + 4x + 1 at the point where x = 3

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When x = 3, we have y = (3^(2) + 4 xx 3 +1) = 22 <br> So, the point of contact is (3, 22) <br> Now y = x^(2) + 4x + 1 rArr (dy)/(dx) = 2x + 4 rArr ((dy)/(dx))_((3, 22)) = (2 xx 3 + 4) = 10 <br> Equation of the tangent is (y - 22)/(x -3) = (-1)/(10) rArr x + 10y - 278 = 0 <br> And, equation of the normal is (y - 22)/(x -3) = (-1)/(10) rArr x + 10y - 223 = 0
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