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Find equation of normal to the curve `y = |x^2-|x||` at `x =-2` In the neighborhood of `x =-2, y=x^2+x`. Hence the point of contact is (-2, 2). |
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Answer» In the neighborhood of `x=-2 ,y =x^(2) +x.` Hence the point of contact is `(-2,2)` `(dy)/(dx)= 2x+1 " "rArr " "(dy)/(dx)|_(x=-2) =-3` So the slope of normal at `(-2,2)` is Hence equation of normal is `(1)/(3) (x+2) =y-2 " "rArr" "3y =x+8` |
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