Thread: Parametric Equation Question

1. Parametric Equation Question

Hi everyone!

I'm new here, and I'll like to ask for help for the following questions. Could someone please provide some general guidelines and hints on how to solve these questions but not give an answer to the questions themselves?

Q1: A curve has parametric equations x=2t2, y=4t. Find the equation of the chord joining the points on the curve where t=p and t=q. I have no idea how to even start solving this. I know that the Cartesian equation of the curve is y2=8x. I have tried substituting in p into x=2t2 and q into y=4t, but that method is wrong.

Q2: This question is similar. Find the equations of the chord AB where at A (t=t1) and at B (t=t2). AB is on a curve with parametric equations x=ct and y=c/t.
I thought that since at AB the variable is in t, I would eliminate the variable c. Therefore, I came up with yt2=x. I found that with this equation, at A, x=y(t1)2, y=((t1)2)/x, and at B, x=y(t2)2, y=((t2)2​)/x. Using these two coordinates, the gradient will be 1/(xy). However, this can never get the answer, which is t1t2y+x=c(t1+t2). How do I get the c inside the answer? Or is it because I'm doing something wrong?

Any help is appreciated!

2. Re: Parametric Equation Question

Why would that method be wrong. That's exactly what you have to do. Get the (x, y) co-ordinates when t = p and when t = q, use these to get the equation of your line.

3. Re: Parametric Equation Question

Hello, LimpSpider!

It looks like you don't understand parametric equations . . .

Q1: A curve has parametric equations: .$\displaystyle \begin{Bmatrix}x&=&2t^2 \\ y&=&4t \end{Bmatrix}$

Find the equation of the chord joining the points on the curve where $\displaystyle t=p$ and $\displaystyle t=q.$

I have tried substituting in p into x=2t2 and q into y=4t, but that method is wrong. . Of course!

You have: .$\displaystyle \begin{Bmatrix}x &=& 2p^2 \\ y &=& 4q\end{Bmatrix} \quad\Rightarrow\quad \text{One point: }\:(2p^2,\,4q)$

So where is the chord?

$\displaystyle \begin{array}{ccccc}\text{When }t = p: & \begin{Bmatrix}x&=& 2p^2 \\ y &=& 4p\end{Bmatrix} & \Rightarrow & (2p^2,4p) \\ \\ \text{When }t=q: & \begin{Bmatrix}x&=& 2q^2 \\ y &=& 4q\end{Bmatrix} &\Rightarrow & (2q^2,4q) \end{array}$

And there are the two points . . .

5. Re: Parametric Equation Question

Hey everyone! Thanks a bunch again for all the answers. If I understand correctly, I must substitute t=variable into BOTH the x and y parametric equations to get the coordinate of one point. I was wrongly thinking all this time that p is a substitute just for x and q for y. . After getting the coordinates, I'm sure I'll be able to get the equations. Thanks again!

(Edit: I have indeed arrived at all the correct answers!! )