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Find the equation of normal to the parabola `y=x^2-x-1`which has equal intercept on the axes. Also find the point where this normalmeets the curve again. |
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Answer» Here, given equation of parabola is not standard. So, we use differentiation method to find the equation of normal. Since normal has equal intercepts on axes, its slope is -1 Now, differentiating equation `y=x^(2)-x-1` on both sides w.r.t. x, we get `(dy)/(dx)=2x-1`. This is the slope of tangent to the curve at any point on it. Thus, slope of normal at any point on it is, `(dx)/(dy)=(1)/(1-2x)=-` (given) `:." "x=1` Putting x=1 in the equation of curve, we get y=-1. So, equation of normal at point (1,-1) on the curve is `y-(-1)=-1(x-1)` `orx+y=0` Solving this equation of normal with the equation of parabola, we have `-x=x^(2)-x-1` `orx^(2)=1` `orx=pm1`. Hence, normal meet parabola again at point whose abscissa is -1. Putting x=-1 in the equation of curve, we get y=1. So, normal meets the parabola again at (-1,1). |
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