Maths help...Parametric Parabola (1 Viewer)

[studentcheese]

I dunno if this is the right forum for Maths problems.. but anyway. I have a simple Maths question that I need help in.

The points P (2ap,ap^2), Q(2aq, aq^2) and T (2at, at^2) are points on the parabola x^2 = 4ay. Find the relation between p,q and t if the chord PQ is perpendicular to the normal at T.

Thanks in advance

 

[xMrRand0m]

I think there's a thread just for Mathematics, but anyway, here's what I think is the solution. Since it's perpendicular, you have to play around with the gradients. So...

x^2=4ay
y=x^2/4a
dy/dx=x/2a
therefore the gradient of tangent at T = t, so gradient of normal at T = -1/t
Gradient of PQ = (y2-y1)/(x2-x1)
= (aq^2-ap^2)/(2aq-2ap)
= a(q-p)(q+p)/2a(q-p)
= (p+q)/2
b/c perpendicular, therefore gradient of PQ x gradient of normal at T = -1
Therefore (p+q)/2 x -1/t = -1
-(p+q)=-2t
p+q=2t

That's what I had, correct me if I'm wrong. =) Hope it's right.

 

[studentcheese]

I think there's a thread just for Mathematics, but anyway, here's what I think is the solution. Since it's perpendicular, you have to play around with the gradients. So...

x^2=4ay
y=x^2/4a
dy/dx=x/2a
therefore the gradient of tangent at T = t, so gradient of normal at T = -1/t
Gradient of PQ = (y2-y1)/(x2-x1)
= (aq^2-ap^2)/(2aq-2ap)
= a(q-p)(q+p)/2a(q-p)
= (p+q)/2
b/c perpendicular, therefore gradient of PQ x gradient of normal at T = -1
Therefore (p+q)/2 x -1/t = -1
-(p+q)=-2t
p+q=2t

That's what I had, correct me if I'm wrong. =) Hope it's right.

Thank you. It's right