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1. Parabola tangents and normals FP1
3)

The point P(4,8) lies on the parabola with equation Find:

a) The value of a

Completed this and is correct a is 4 so the equation is

b) an equation of the normal to C at P

Answer: y = -x + 12

The normal to C at P cuts the parabola again at the point Q. Find

c) the coordinates of Q

so i subbed the normal to the parabola equation and heres my working:

Now heres my problem, If I sub the x value of 36 into y^2 = 16x then I will get 24 and -24, and If I sub it into the normal equation of y = -x + 12 I will get -24. Which one am I supposed to use and how do I know which one to use?

thanks.
2. Re: Parabola tangents and normals FP1
(Original post by lemonpwns1)
...
It crosses c again at (36,-24) as that point lies on both the normal and on c, the point (36,24) lies on C but not on the normal so it clearly can't be the point where they cross. You could just draw a sketch if you were unsure which would help a lot.

EDIT: Also it made me happy to see you completed the square, far too many people use the formula
Last edited by TheJ0ker; 02-06-2012 at 19:13.
3. Re: Parabola tangents and normals FP1
(Original post by TheJ0ker)
It crosses c again at (36,-24) as that point lies on both the normal and on c, the point (36,24) lies on C but not on the normal so it clearly can't be the point where they cross. You could just draw a sketch if you were unsure which would help a lot.

EDIT: Also it made me happy to see you completed the square, far too many people use the formula
Completing the square is much quicker - Thank you.
4. Re: Parabola tangents and normals FP1
(Original post by lemonpwns1)
Completing the square is much quicker - Thank you.
If you've studied roots of polynomials you could notice:

After substuting for y from the eqn of the normal into that of the parabola, and rearranging to get a quadratic.

Then the x co-ordinate of P is one of the values satisfying that quadratic, i.e. one of the roots is 4. And since the sum of the roots is 40, the other solution must be x=36. So you don't actually have to solve the quadratic.
Last edited by ghostwalker; 02-06-2012 at 20:56.
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