Real Madrid said:
The focal chord that cuts the parabola x^2=-6y at (6,-6) cuts the parabola again at X. Find the coordinates at X.
Show all working.
ok im not gunna do it coz im about to go out
but ill push you in the right direction
find the equation of the focal chord, noting the focus is (0,-3/2) y finding the gradient (m=y2-y1/x2-x1) and using the point gradient formula (y-y1=m(x-x1) )
then when you have it as an expression such as y=
12x +3(not the focal chord, just random equation)
sub in the underlined bit into the parabola
then solve for x, should get x=6 and some other value
the sub the other value into either equation for y