# Thread: Exercise 11.10 (Locus Problems) - Question 1

1. ## Exercise 11.10 (Locus Problems) - Question 1

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!

2. ## Re: Exercise 11.10 (Locus Problems) - Question 1

Originally Posted by 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!
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)

3. ## Re: Exercise 11.10 (Locus Problems) - Question 1

Originally Posted by Drongoski
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?

4. ## Re: Exercise 11.10 (Locus Problems) - Question 1

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.

5. ## Re: Exercise 11.10 (Locus Problems) - Question 1

Originally Posted by 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.
Okay - thanks!

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