Answer
$(1.5,1.25)$
Work Step by Step
The equation of parabola is
$$y=x^2-1$$
Taking derivative
$$\begin{align*}
\frac{dy}{dx}& =\frac{d(x^2-1)}{dy} \\
&=\frac{d(x^2)}{dy}-\frac{d(1)}{dy} \\
& =2x-0 \\
& =2x
\end{align*}$$
At $ x=-1$
$$ \frac{dy}{dx} =2(-1)=-2 =m, $$
where m is the slope of tangent at the point (-1,0).
Let $m_1$ be the slope of the normal at the same point. Then
$$m m_1=-1$$
Putting $m=-2$ in the above equation
$$-2m_1=-1$$
$$m_1=\frac{1}{2}$$
$m_1$ is the slope of normal through the point $(-1,0)$
The general equation of normal through the point $(x_1,y_1)$ with slope $m_1$.
$$ y-y_1=m_1(x-x_1)$$
Putting $(x_1,y_1)=(-1,0)$ and $m_1=\frac{1}{2}$ in the above equation
$$y-0=\frac{1}{2}(x+1)$$
$$y=\frac{1}{2}(x+1)$$
$$2y=x+1$$
$$2y-1=x$$
$$x=2y-1$$
Putting $x=2y-1$ in $y=x^2-1$
$$y=(2y-1)^2-1$$
$$y=4y^2-4y+1-1$$
$$y=4y^2-4y$$
$$0=4y^2-4y-y$$
$$0=4y^2-5y$$
$$4y^2-5y=0$$
$$y(4y-5)=0$$
This imply that either
$$y=o$$
Or
$$4y-5=0$$
Putting $y=0$ in equation $x=2y-1$,we have $x=-1$
$(-1,0)$ is the given point
At
$$4y-5=0$$
$$4y=5$$
$$y=\frac{5}{4}=1.25$$
Putting $y=1.25$ in equation $x=2y-1$
$$x=2(1.25)-1=1.5$$
$(1.5,1.25)$ is the other point where normal intersect parabola.
