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