Exercise 11.10 (Locus Problems) - Question 1 (1 Viewer)

[Heresy]

1a) Find the equation of the focal chord PF on the parabola x^2 = 8y where P= (-8,8) and F is the focus.

I have already found that the equation is 3x +4y - 8 = 0


b) Find the coordinates of Q where the focal chord intersects the parabola again.

THIS IS THE MAIN QUESTION I'M HAVING TROUBLE WITH.


Any help would be greatly appreciated!

 

[Drongoski]

1a) Find the equation of the focal chord PF on the parabola x^2 = 8y where P= (-8,8) and F is the focus.

I have already found that the equation is 3x +4y - 8 = 0


b) Find the coordinates of Q where the focal chord intersects the parabola again.

THIS IS THE MAIN QUESTION I'M HAVING TROUBLE WITH.


Any help would be greatly appreciated!

You solve the 2 equations simultaneously.

From parabola you get: 8y = x²

From eqn of chord: 8y = 2(8-3x) = 16-6x

.: x² = 16 - 6x ==> x² + 6x -16= 0 ==> (x+8)(x-2) = 0

So chord intersects parabola at x = -2 as well. x=2 ==> y = (8-3x)/4 = (8 - 3*2)/4 = 1/2 = 2

So Q has co-ords (2,1/2)

 

[Heresy]

You solve the 2 equations simultaneously.

Thanks. I've solved it but I have a question - when we factor the quadratic we get and we recieve 2 solutions; without checking the answers - how do I know which x - variable to sub in?

 

[Drongoski]

One of the end-points of the chord has x = 8, one of the 2 solutions of the quadratic eqn; so Q corresponds to the other solution x = 2.

 

[Heresy]

One of the end-points of the chord has x = 8, one of the 2 solutions of the quadratic eqn; so Q corresponds to the other solution x = 2.

Okay - thanks!