Locus, parabola question (1 Viewer)

[currysauce]

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

 

[withoutaface]

well first you find the coordinates of the focus:

x²=4ay

:. 4a=-6
a=-3/2

focus (0, -3/2)

Find the equation of the line between that point and the focus:

m= (-6+3/2)/(6)
=-9/12
=-3/4

(-3/4)x=y+3/2

eqate this to the equation of the parabola

(-3/4)x-3/2=x²/-6

then expand it out, simplify and use the quadratic formula to solve for the x point.

If you need me to do the last part just post again, but I think its pretty straightfoward

 

[currysauce]

umm

isn't it (0, -3/2) for the focus?

 

[withoutaface]

Yeah, whoops

Fixed now.

 

[currysauce]

i still don't know how to do it... trying to solve it with the formula gives a negative root

could ya show me how to fix?

 

[withoutaface]

The guy isn't doing 4u, he's in year 11 planning to do 4u, so lets learn to read before criticising someone shall we?


And by negative root do you mean a negative discriminant or just a negative root?

Cause if its a negative root its correct.