1. ## Locus/Hyperbola question

Hey guys,

I'm really bad at these types of questions, I don't know what it is about them but they always seem to stump me. Here it is,

The variable point $P(a\sec t, b\tan t)$ is on the hyperbola with equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and N is the point (3a, 3b). The point Q lies on the line PN so that PQ = 2QN. As t varies find, in cartesian form, the locus of Q.

I tried to find the line PN by finding its gradient by

$\frac{3b - b\tan t}{3a - a\sec t}$

and then using the usual method for finding the equation of a straight line, that is

$\frac{y-3b}{x-3a} = \frac{3b - b\tan t}{3a - a\sec t}$

but then I get stuck and cannot see what to do next. Do I need to find the point Q? I know I need to eliminate t, but I just can't see how to approach this one.

Are there any general tips and tricks common to conic section questions like this? Given a formula I can identify which conic it is, and its directrix, foci etc. But when the questions get slightly harder (like this one) I just go to pieces.

Any help would be much appreciated

Stonehambey

2. Hi

When you say that the point Q lies on the line PN so that PQ = 2QN, do you mean that Q is between P and N ? Otherwise there are 2 such points

P------------Q1-------N-------------------Q2

3. Originally Posted by running-gag
Hi

When you say that the point Q lies on the line PN so that PQ = 2QN, do you mean that Q is between P and N ? Otherwise there are 2 such points

P------------Q1-------N-------------------Q2
"The point Q lies on the line PN so that $\vec {PQ} = 2\vec{QN}$"

That is the quoted from the question verbatim

4. OK

Actually you don't need to find a Cartesian equation of line NP

From $\overrightarrow{PQ} = 2\:\overrightarrow{QN}$ you can find

$\overrightarrow{NQ} = \frac{1}{3}\:\overrightarrow{NP}$

from which you can find the coordinates of Q and then eliminate t

Spoiler:

$\frac{9(x_Q-2a)^2}{a^2} - \frac{9(y_Q-2b)^2}{b^2} = 1$

$\frac{(x_Q-2a)^2}{\left(\frac{a}{3}\right)^2} - \frac{(y_Q-2b)^2}{\left(\frac{b}{3}\right)^2} = 1$

5. ok thanks, I will give this question another go when I get back from work this evening

6. I'm afraid I'm still having trouble with this one

Do I need to find the lengths of NQ and NP using the pythagorean theorem? Then by setting NQ = (1/3)NP will eliminate t?

Thanks again

7. Sorry, could you better explain where you get stuck ?

8. Originally Posted by running-gag
Sorry, could you better explain where you get stuck ?
Sure,

Well as you say we know that $\overrightarrow{NQ} = \frac{1}{3}\:\overrightarrow{NP}
$

So does this mean I just have to work out the lengths of $\overrightarrow{NP}$ and $\overrightarrow{NQ}$ and continue the problem that way? I really don't see the way forward and I tend to write out pages of algebra on these types of problems, so I just wanna know I'm going in the right direction

I'm sorry, I have a whole host of conic section problems I can't do. It seems to be a complete brick wall for me. I just can't seem to answer questions on them no matter how much I practice

9. ## Ratios

Hello Stonehambey
Originally Posted by Stonehambey
Sure,

Well as you say we know that $\overrightarrow{NQ} = \frac{1}{3}\:\overrightarrow{NP}
$

So does this mean I just have to work out the lengths of $\overrightarrow{NP}$ and $\overrightarrow{NQ}$ and continue the problem that way?
No. What you need is this:

If $A, B$ have position vectors $\vec{a}, \vec{b}$ then the position vector $\vec{r}$ of the point $R$ that divides $AB$ internally in the ratio $\lambda:\mu$ is given by:

$\vec{r} = \frac{\mu\vec{a}+\lambda\vec{b}}{\lambda+\mu}$

This is a very useful formula. Learn it!

The point P has coordinates $(a\sec t, b\tan t)$; the point $N$ has coordinates $(3a, 3b)$, and $Q$ divides $PN$ internally in the ratio $2:1$. (That's what $\vec{PQ} = 2\vec{QN}$ means.) So, using the formula above, the coordinates of $Q$ are:

$\Big(\tfrac13(a\sec t + 6a), \tfrac13(b\tan t + 6b)\Big)$

So the locus of Q is given by

$x =\tfrac13(a\sec t + 6a), y=\tfrac13(b\tan t + 6b)$

i.e. $\sec t = \frac{3(x-2a)}{a}, \tan t =\frac{3(y-2b)}{b}$

Eliminating $t$:

$\frac{9(x-2a)^2}{a^2}-\frac{9(y-2b)^2}{b^2}=1$