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Find the equations of the tangent and the normal to the curve `y (x -2) (x -3) - x + 7 = 0` at the point where it cuts the x-axis |
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Answer» The curve cuts the x-axis, where `y = 0` Putting y = 0 in the given equation, we get `x = 7` `:.` the point of contact is (7, 0) Now, `y (x -2) (x -3) - x + 7 = 0 rArr y(x^(2) - 5x + 6) - x + 7 = 0` `rArr (x^(2) - 5x + 6).(dy)/(dx) + y (2x -5) - 1 = 0 rArr (dy)/(dx) = (1 + 5y - 2xy)/(x^(2) - 5x + 6)` `:. ((dy)/(dx))_((7","0))= ((1 + 5 xx 0 - 2 xx 7 xx 0)/(49 - 35 + 6)) = (1)/(20)` So, the equation of the tangent is `(y - 0)/(x -7) = (1)/(20) or x - 20y - 7 = 0` Also, the equation of the normal is `(y - 0)/(x -7) = - 20 or 20x + y - 140 = 0` |
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