Focal chord that cuts the parabola x²= -6y at (6,-6) cuts the parabola again at X. Find the coordinates of X ?

Answer 1

First find the coordinates of the vertex and focus of the parabola.

-6y = x²

The vertex (h, k) = (0,0).

4p = -6
p = -6/4 = -3/2

The focus is
(h, k + p) = (0, 0 - 3/2) = (0, -3/2).

So the chord is the line that passes thru the points
(6, -6) and (0, -3/2).

Calculate the slope.

m = ∆y/∆x = (-3/2 + 6) / (0 - 6) = (9/2) / (-6) = -9/12 = -3/4

The equation of the line is:
y + 6 = (-3/4)(x - 6) = (-3/4)x + 9/2
y = (-3/4)x - 3/2

Find the intersection of the parabola and the line.

-6y = x²

y = (-1/6)x² = (-3/4)x - 3/2
2x² = 9x + 18
2x² - 9x - 18 = 0
(2x + 3)(x - 6) = 0
x = -3/2, 6

Solve for y.

y = (-1/6)x² = (-1/6)(-3/2)² = (-1/6)(9/4) = -9/24 = -3/8

The other point is X(-3/2, -3/8).

Answer 2

x^2 = - 6 y . . with line passing (6, -6)
parabola is facing down, vertex at (0,)
it is symmetrical to y -axis

line will pass also at (-6, -6)