maths problem (1 Viewer)

[Real Madrid]

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.

 

[tacogym27101990]

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

 

[youngminii]

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.

So in x^2=-6y, the Focus would be (0,-1/2)
And since the Chord goes through the Focus, we have two points with which we find the Gradient
Therefore m=-11/12

Using Point-Gradient formula, 11x+12y+6=0 is the equation of the Focal Chord
Solving Simultaneously to find the point of intersections between the Parabola and the Focal Chord
We end up with x=-1/2 or x=6
When x=6, y=-6(the Point we were given to start with)
When x=-1/2, y=-1/24
Therefore the point X is (-1/2, -1/24)
Ta-da

 

[Danneo]

Lol mate you got the wrong answer though i assume your method was correct, m=-3/4, just sub y=(x^2)/-6 into the tangent equation