Question 1093704: The focal chord that cuts the parabola x^2= -6y at (6, -6) and cuts the parabola again at X. Find the coordinates of X.
I tried to solve it...what am I doing wrong:
concave down
-4a=-6
a= -3/2
vertex at (0,0)
so, focus is at (0,3/2)
I used the 2 point formula y - y1 = m(x - x1) and found the gradient of (6,-6) and (0,3/2):
m= -4/3
y-6=-4/3 (x+6)
y=-4/3x+14
used simultaneous equations x^2= -6y and y=-4/3x+14 to find the intersection points but did not get the right answer.. is my process right?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! any chord that passes through the Parabola's focus point is a focal cord
:
x^2 = -6y
:
we are given one intersection point of the focal cord, namely (6, -6)
:
the center is at (0,0)
:
4p = -6
:
p = -6/4 = -3/2 = -1.5
:
the focus is -1.5 units from y coordinate of the center, namely (0, -1.5)
:
we have two points on the cord, (6, -6) and (0, -1.5)
:
slope = (-1.5 -(-6)) / (0 -6) = 4.5 / -6 = -0.75 = -3/4
:
y = -3x/4 +b
:
use point (6, -6) to find b
:
-6 = -3(6)/4 + b
:
-24 = -18 +4b
:
4b = -6
:
b = -6/4 = -1.5
:
y = -3x/4 -1.5
:
-3x/4 -1.5 = -x^2/6
:
-9x -18 = -2x^2
:
2x^2 -9x -18 = 0
:
(2x+3) * (x-6) = 0
:
x = -3/2 and x = 6
:
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we choose x = -1.5 then
:
y = -3(-1.5)/4 -1.5 = 1.125 -1.5 = −0.375
:
coordinates of X are (-1.5, -0.375)
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