Parametric equation? extension question?

Question

Find the equation of the locus of the midpoint M of all chords PQ where P(2ap,ap^2) and Q (2aq,aq^2) lie on the parabola x^2=4ay and PQ passes through (0,2a). I know the eqn of the locus will end up as a parabola too, but how do I work it out?

So far, I know vertex = (0,2a) and midpoint of PQ (ap+aq, (ap^2+aq^2)/2)

Answer 1

Okay, here is how you begin these type of questions.

Firstly, you need to find a relationship with p's and q's so you can eliminate the parameters and find a cartesian equation for the locus.

We begin by finding the equation of the chord PQ.

First find the gradient:

Using the gradient formula:

m = [ap^2 - aq^2] / [2ap - 2aq]

Factorising using the difference of two squares:

= a(p - q)(p + q) / 2a(p - q)

Cancel the a's and the (p - q):

m = (p + q) / 2

Using the point gradient equation to find the equation:

y - ap^2 = [(p + q) / 2] [x - 2ap]

Taking the 2 to the other side and expanding:

2y - 2ap^2 = (p + q)x - 2ap^2 - 2apq

Cancelling the 2ap^2 and dividing by 2.

y = [(p + q) / 2]x - apq

Okay now that we have the equation, we know that it has to pass through the vertex (0,2a). Now we substitute x = 0 and y = 2a into that equation.

2a = -apq

Therefore pq = -2. (cancelling a and changing sign)

This is the relationship we need to find the cartesian equation. Now we find the midpoint like you did, i won't go through it because it is really simple.

We get this:

(ap+aq, (ap^2+aq^2)/2)

Okay now write it like:

x = a(p + q)   and y = a(p^2 + q^2) / 2.

We need to eliminate the p's and q's from these by substituting and rearranging etc. Notice how we found pq = -2 from before, since we know this must be used, you have to use a special relationship involving that. This relationship is:

Since: (p + q)^2 = p^2 + 2pq + q^2

p^2 + q^2 = (p + q)^2 - 2pq.

Notice how there is a p^2 + q^2 in the y equation, and that there is a pq which we know the value of.

p^2 + q^2 = (p + q)^2 + 4 since pq=-2.

Now to remove the (p + q)^2.

Notice the similarity with the equation involving x.

x = a(p + q)

x/a = p + q

x^2 / a^2 = (p + q)^2

Therefore

p^2 + q^2 = x^2 / a^2 + 4

or

p^2 + q^2 = [x^2 + 4a^2] / [a^2]

Now we sub it into y to find the final equation.

y = a(p^2 + q^2) / 2.

y = a([x^2 + 4a^2] / [a^2]) / 2.

Cancel the a's:



y = [x^2 + 4a^2] / [a] / 2.

Therefore the final cartesian equation is:

y = (x^2) / (2a) + 2a

Notice the removal of the p's and q's, and how the a's will make the equation vary as the parabola changes. They could ask you to do this but with a real equation of a parabola like x^2 = 2y, and this would just mean the equation would no longer have a's (y = x^2).

Answer 2

What class is this for?

Slope of the chord:

m = (aq^2 - ap^2)/(2aq - 2ap) = (q^2 - p^2)/(2q - 2p) = (q + p)/2

Must go through the point (0,2a) so:

y = mx + b

2a = b

y = [(q + p)/2]x + 2a is the equation of the chord

But the fact that it must go thorugh the point (0,2a) sets constraints on the values of p and q. The line must also go through (2ap,ap^2) so:

ap^2 = [(q + p)/2](2ap) + b

b = -pqa

Then 2a = -pqa or pq = -2

Now solve the chord equation for q + p:

(q + p) = 2(y - 2a)/x

Now this x can be set equal to the midpoint value:

x = a(p + q) = a2(y - 2a)/x

x^2 = 2ay - 4a^2

And this is the equation you are looking for.

Check by putting in x from the midpoint:

(ap + aq)^2 = 2ay - 4a^2

a^2p^2 + 2a^2qpq + a^2q^2 = 2ay - 4a^2

ap^2 + 2apq + aq^2 = 2y - 4a

Now use the midpoint y to replace ap^2 + aq^2:

2apq + 2y = 2y - 4a

2apq = -4a

And remember that pq = -2 from the above discussion:

2a(-2) = -4a

And it seems to  be OK.

The equation is:

x^2 = 2ay - 4a^2