Question

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

Answer 1

answer is in the picture.

Answer 2

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).