# Parabola Help

1. Sep 8, 2007

### Mattofix

[SOLVED] Parabola Help

1. The problem statement, all variables and given/known data

Click me

2. Relevant equations

Unknown

3. The attempt at a solution

I can complete a) fine, by differentiating and sticking in y value to get m for tangent then -1/m for normal (-p). Then i used (y - y1) = m (x - x1) to provide normal equation.

I have tried a few ways for b) but i always end up the normal equation - no good. The only method which gives me q in terms of p is if i make the normal equation equal the parabola equation then solve - the answer i get doesn’t look right at all, and it doesn’t leave me in a good position for c)

is there something i should know about parabola’s to be able to solve b) or is it a route i haven’t seen?

Any help would be much appreciated

2. Sep 8, 2007

### rock.freak667

Well the normal passes through Q, so you could sub the points of Q into the normal and q in terms of p

3. Sep 8, 2007

### Mattofix

How would you arrange 2q + pq^2 = 2p + p^3 ?

4. Sep 8, 2007

### rock.freak667

Well normally I'd leave it like that...but I am not sure really how q in terms of p is meant as..

5. Sep 9, 2007

### Mattofix

it means q = 3p or something like that, and you need it looking like that to be able to do (c), im really think im missing something

6. Sep 9, 2007

### HallsofIvy

Staff Emeritus
$2q + pq^2 = 2p + p^3$ is a quadratic in q. Rewrite it in the form $pq^2+ 2q- (2p+p^3)= 0$ and solve that equation for q, using the quadratic formula, leaving p in the formula. That's what "in terms of p" means

7. Sep 9, 2007

### Mattofix

Great, could you confirm q = -2/p - p ?

8. Sep 9, 2007

### HallsofIvy

Staff Emeritus
There are, of course, two solutions. q= -2/p- p is one. The other is obvious and easy to check in the original equation.

9. Sep 9, 2007

### arildno

Well, simplify:
$$q=\frac{-2\pm\sqrt{4+4p(2p+p^{3})}}{2p}=\frac{-1\pm\sqrt{1+2p^{2}+p^{4}}}{p}=\frac{-1\pm(1+p^{2})}{p}$$

10. Sep 12, 2007

thanks guys