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 normal meets the curve a

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).