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Find the equation of normal to parabola `y=x^(2)-3x-4` (a) at point (3,-4) (b) having slope 5. |
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Answer» Correct Answer - (a) `3x-y-13=0`, (b) `5x-y-20=0` We have parabola `y=x^(2)-3x-4` Differentiating w.r.t. x, we get `(dy)/(dx)=2x-3` (a) `((dy)/(dx))_((2","-4))=3` So, equation of normal is `4+4=3(x-3)` `or" "3x-y-13=0` (b) `(dy)/(dx)=2x-3=5` `:." "x=4` So, y=16-12-4=0. Therefore, equation of normal is `y-0=5(x-4)` `or" "5x-y-20=0` |
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