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Parametric equations of curves

I can do (a) and I can do most of (b)- I can get the distances of P to A (PA) and P to B (PB) but then I don't know what to do with them to show PB=2PA??? Thanks!

Original post by rpnom

I can do (a) and I can do most of (b)- I can get the distances of P to A (PA) and P to B (PB) but then I don't know what to do with them to show PB=2PA??? Thanks!

Well, if you know

PA

and

PB

then you just need to say that the first is twice the second? Can we see your working out?

OP

Original post by Zacken

Well, if you know

PA

and

PB

then you just need to say that the first is twice the second? Can we see your working out?

Original post by rpnom

...

You've said

A

is the point

\left(\frac{3}{2p}, 0\right)

and

B

is the point

(0, 3p^2)

. I agree with this.

Now the point P is

\left(\frac{1}{p}, p^2\right)

. Can you now find the correct distances

PA

and

PB

using the standard distance formula?

OP

Original post by Zacken

You've said

A

is the point

\left(\frac{3}{2p}, 0\right)

and

B

is the point

(0, 3p^2)

. I agree with this.

Now the point P is

\left(\frac{1}{p}, p^2\right)

. Can you now find the correct distances

PA

and

PB

using the standard distance formula?

I sort of did that didn't I? I found PA^2 and PB^2 using that formula? :/

Original post by rpnom

I sort of did that didn't I? I found PA^2 and PB^2 using that formula? :/

Look at your PA^2, why are you subtracting 0 from both of them?

It should be

PA^2 = \left(\frac{3}{2p} - \frac{1}{p}, 0 - \frac{1}{p^2}\right)

and same for PB - you seem to be treating

P

as

(0,0)

.

Original post by rpnom

Despite the fact you subtracted 0 twice (as Zacken pointed out), you got the correct distances. Now pull out a factor of 4 in your

PB^{2} = \frac{1}{p^{2}}+4p^{4}

and you should find that you get

PB^{2} = 4PA^{2}

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