please help, parametric parabola question (1 Viewer)

[Schniz]

this question is making me so mad


find the equation of the chord joining the points t = 3 and t = -1/3 on the parabola x^2 = 4ay


thanks

 

[Riviet]

I am assuming your standard parametric equations are P(2at,at²). Simply substitute t=3 and t=-1/3 into P to obtain two points, find the gradient using m=(y2-y1)/(x2-x1) and use the point-gradient formula to obtain the equation of the chord. Hope that helps.

 

[SoulSearcher]

We know that the standard form of the parametric equations of a parabola is equal to:
x = 2at
y = at²
Substitute both values of t into the above 2 equations
t = 3, x = 6a and y = 9a; i.e. (6a, 9a)
T = -1/3, x = -2a/3 and y = a/9; i.e. (-2a/3, a/9)
From here, you can use those 2 points to find the gradient of the line joining those 2 points, and then find out the equation of the chord joining those 2 points.

 

[echelon4]

been a while since i did parametrics. Here goes....

Sub the t's into x=2at, y=at^2

the two points are:

P(6a, 9a)
Q(-2a/3, a/9)

Use the two points equation.

(x-x1)/(y-y1)=(x2-x1)/(y2-y1)

comes out to be 4x-3y+3=0 i think......

 

[SoulSearcher]

You're right, echelon4