`show that the normal to the parabola y²=8x at (2 4)meets the parabola again in(18, -12)`

Question

asked Jul 19, 2022 54.9k views

2 Answers

MooCow · Answer 1 · 2022-07-23T19:52:39+0000

Answer:

See Below.

Explanation:

We want to show that the normal line to the parabola:

y^2=8x

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:

$\displaystyle (d)/(dx)\left[ y^2\right]=(d)/(dx)\left[8x\right]$

Implicit differentiation:

$\displaystyle 2y(dy)/(dx)=8$

Therefore:

$\displaystyle (dy)/(dx)=(8)/(2y)=(4)/(y)$

Then the slope of the tangent line to the point (2, 4) is:

$\displaystyle (dy)/(dx)\Big|_(y=4)=(4)/((4))=1$

Thus, the slope of the normal line is -1.

And since it passes through the point (2, 4), by the point-slope form:

y-(4)=-1(x-2)

Simplify:

y=-(x-2)+4

By letting x = 18:

y=-(18-2)+4=-16+4=-12

So, the normal line indeed passes through (18, -12).