Answer
$\left(\frac{3}{2},\frac{5}{4}\right)$
Work Step by Step
Given a parabola $y=x^2-1$.
Find the slope of the tangent line to the parabola at $(-1,0)$: $m_{tangent}=\frac{dy}{dx}|_{x=-1}=(2x)|_{x=-1}=2\cdot (-1)=-2$
Since the normal line and the tangent line are perpendicular, the slope of the normal line to the parabola at $(-1,0)$ is
$m_{normal}=\frac{-1}{m_{tangent}}=\frac{-1}{-2}=\frac{1}{2}$
Find the equation of the normal line:
$y-0=m_{normal}(x-(-1))$
$y=\frac{1}{2}(x+1)$
Find the intersection points between the normal line and the parabola:
$x^2-1=\frac{1}{2}(x+1)$ (Multiply by 2)
$2x^2-2=x+1$
$2x^2-x-3=0$
$(x+1)(2x-3)=0$
$x=-1\vee x=\frac{3}{2}$
For $x=-1$, $y=(-1)^2-1=1-1=0$
For $x=\frac{3}{2}$, $y=(\frac{3}{2})^2-1=\frac{9}{4}-\frac{4}{4}=\frac{5}{4}$
Thus, the normal line intersects the parabola for the second time at the point $\left(\frac{3}{2},\frac{5}{4}\right)$.
