Parabola Help [SOLVED] with Tangent and Normal Equations

[Mattofix]

[SOLVED] Parabola Help

Homework Statement

Click me

Homework Equations

Unknown

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

 

[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

 

[Mattofix]

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

 

[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..

 

[Mattofix]

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

 

[HallsofIvy]

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

 

[Mattofix]

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

 

[HallsofIvy]

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

 

[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}

 

[Mattofix]

thanks guys