Answer:
See Below.
Step-by-step explanation:
We want to show that the normal line to the parabola:
[tex]y^2=8x[/tex]
At the point (2, 4) meets the parabola again at (18, -12).
First, find the tangent line to the parabola at the point (2, 4). We can take the derivative of both sides with respect to x:
[tex]\displaystyle \frac{d}{dx}\left[ y^2\right]=\frac{d}{dx}\left[8x\right][/tex]
Implicit differentiation:
[tex]\displaystyle 2y\frac{dy}{dx}=8[/tex]
Therefore:
[tex]\displaystyle \frac{dy}{dx}=\frac{8}{2y}=\frac{4}{y}[/tex]
Then the slope of the tangent line to the point (2, 4) is:
[tex]\displaystyle \frac{dy}{dx}\Big|_{y=4}=\frac{4}{(4)}=1[/tex]
Thus, the slope of the normal line is -1.
And since it passes through the point (2, 4), by the point-slope form:
[tex]y-(4)=-1(x-2)[/tex]
Simplify:
[tex]y=-(x-2)+4[/tex]
By letting x = 18:
[tex]y=-(18-2)+4=-16+4=-12[/tex]
So, the normal line indeed passes through (18, -12).
