Question

Question #ed0b6

Answer 1

Two tangents intersect at `(0,2)`.

Explanation:

Given the parabola as `y=ax^2`, it's focal point is at `(0,1/(4a))`.
In our case of `x^2=-8y` (or, equivalently, `y=-1/8x^2`) the focal point is at `(0,-2)`.

graph{-0.125x^2 [-5, 5, -5, 1]}

Therefore, the focal chord is represented by a horizontal line parallel to X-axis and going through point `(0,-2)`.
It crosses the parabola at `x=+-4` since for `y=-2` `x^2=(-8)*(-2)=16`.

Now we can find the slope of a tangent to parabola at points `x=+-4`.
The derivative of `y=-1/8x^2` is `y'=-1/4x`.
At points `x=+-4` it's equal to `+-1`, that is it's at angle `45^o` to the X-axis.

Going through point `(+-4,-2)` at angle `45^o` to the X-axis, two tangents intersect at point `(0,2)`.