Parabola as a locus question (1 Viewer)

[lordinance]

I don't know how to solve this and would be appreciated if someone could do it

The focal chord that cuts the parabola X^2=-6y at (6,-6) cuts the parabola again at X.
Find the coordinates of X.

Answer
X=(-3/2,-3/8)

 

[Revacious]

The focal chord that cuts the parabola X^2=-6y at (6,-6) cuts the parabola again at X.
Find the coordinates of X.

x^2 = -4ay where a is the focal length. This is the form of a concave left parabola, vertex origin. In this case, 4a = 6, a = 3/2

So the focus is at (-3/2, 0)

the focal chord passes through the focus (-3/2,0) and (6,-6) so find the equation of the line by either finding gradient and using y-y1 = m(x-x1) or using 2 point form of a line. Then solve simultaneously with parabola. You should get 2 solutions as with all quadratics, one should be 6 and the other should be -3/2

apologies if i did it wrong; its pretty late here haha.