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 again.

Explanation

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).