1. ## locus - parabola

p127 q8
Find the equation of the locus of the mid-point of a vairable chord thro' the point (a , 2a ) of the parabola $\displaystyle y^2=4ax$
thanks

edit: my working: let p be a point on $\displaystyle y^2=4ax$ , P = $\displaystyle (at^2 , 2at)$.
let A = (a, 2a)
equ of PA: $\displaystyle \frac {y-2a}{x-a} = \frac{2at-2a}{at^2-a}$
$\displaystyle y= \frac {2(x-a)}{t+1} +2a$
have no idea how to do

2. Hello,

Originally Posted by afeasfaerw23231233
p127 q8
Find the equation of the locus of the mid-point of a vairable chord thro' the point (a , 2a ) of the parabola $\displaystyle y^2=4ax$
thanks

edit: my working: let p be a point on $\displaystyle y^2=4ax$ , P = $\displaystyle (at^2 , 2at)$.
let A = (a, 2a)
equ of PA: $\displaystyle \frac {y-2a}{x-a} = \frac{2at-2a}{at^2-a}$
$\displaystyle y= \frac {2(x-a)}{t+1} +2a$
have no idea how to do
Hmmm you're both near to it and not

Let P like you did, and A(a, 2a).

The equation of the line is y=mx+n, but actually, you don't need it (if I read correctly).

The midpoint has its coordinates equal to half the sum of the other coordinates. Let M be the midpoint.

$\displaystyle M \left(\frac{a+at^2}{2} ~,~\frac{2a+2at}{2}\right) ...$

Am I right ??

3. thanks! what a fool i am..
i misread the question. i thought it asked me to find the mid point of the intersection of the chord and the parabola. can i find the locus of the mid point if it is the case?

4. Originally Posted by afeasfaerw23231233
thanks! what a fool i am..
i misread the question. i thought it asked me to find the mid point of the intersection of the chord and the parabola. can i find the locus of the mid point if it is the case?
I've drawn the locus of the midpoints of all chords with A as one endpoint.

If the parabola has the equation

$\displaystyle y^2 = 4ax$

The focus has the coordinates F(a, 0) and the point A(a, 2a) is placed on the parabola. (In my example I've used a = 3)

$\displaystyle M_1$ and $\displaystyle M_2$ are midpoints of the chords.

The locus is a parabola too.

5. thanks. i didn't see that a,2a is always a point on the parabola. i thought it was a point not lying on it. next time i have to sub the value of the point to the equation of the parabola to see if it is a point on it first.