Question #ed0b6

1 Answer
Jul 2, 2016

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)#.