# Coordinate system help

• Mar 24th 2008, 05:23 AM
rednest
Coordinate system help
Question:

The tangents to the ellipse with equation $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$ at the points $P(acost,bsint)$ and $Q(-asint,bcost)$ intersect at the point $R$. As $t$ varies, show that $R$ lies on the curve with equation $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 2$.
• Mar 24th 2008, 05:41 AM
mr fantastic
Quote:

Originally Posted by rednest
Question:

The tangents to the ellipse with equation $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$ at the points $P(acost,bsint)$ and $Q(-asint,bcost)$ intersect at the point $R$. As $t$ varies, show that $R$ lies on the curve with equation $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 2$.

I don't have the patience to do the details, but here's an outline of an approach:

You need to get the coordinates x = x(t) and y = y(t) of the intersection point R of the two tangents. The coordinates of R obviously define a curve parametrically - hopefully the curve $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 2$ !

To get the intersection point you need the equation of the two tangents. To get these equations, you need the gradient and a known point. The known point is given in each case. You get the gradient from $\frac{dy}{dx}$.

I'd use implicit differentiation to get $\frac{dy}{dx} = - \frac{b^2 x}{a^2 y}$.

At the point $P(a \cos t, \, b \sin t)$, $\frac{dy}{dx} = - \frac{b \cos t}{a \sin t}$.

So the tangent at this point has equation

$y = - \frac{b \cos t}{a \sin t} (x - a \cos t) + b \sin t$ .... (1)

At the point $Q(-a \sin t, \, b \cos t)$, $\frac{dy}{dx} = \frac{b \sin t}{a \cos t}$.

So the tangent at this point has equation

$y = \frac{b \sin t}{a \cos t} (x + a \sin t) + b \cos t$ .... (2)

Solve equations (1) and (2) simultaneously to get the intersection point R.

I reserve the right for the above outline to contain careless errors and typos.
• Mar 24th 2008, 06:57 AM
rednest
Next part?
I have managed to get upto point where I need to solve simultaneously myself, but I can't get to the final answer. Could you perhaps show me how it's done? Thanks.
• Mar 24th 2008, 06:41 PM
mr fantastic
Quote:

Originally Posted by rednest
I have managed to get upto point where I need to solve simultaneously myself, but I can't get to the final answer. Could you perhaps show me how it's done? Thanks.

Well, it would have saved time to know what point you got up to and where you got stuck.

Can you post the working you've done in tying to solve simultaneously.
• Mar 24th 2008, 07:05 PM
rednest
Quote:

Originally Posted by mr fantastic
Well, it would have saved time to know what point you got up to and where you got stuck.

Can you post the working you've done in tying to solve simultaneously.

Since
$y = - \frac{b \cos t}{a \sin t} (x - a \cos t) + b \sin t$
$y = \frac{b \sin t}{a \cos t} (x + a \sin t) + b \cos t$
I let $\frac{b \sin t}{a \cos t} (x + a \sin t) + b \cos t = - \frac{b \cos t}{a \sin t} (x - a \cos t) + b \sin t$.

But I'm stuck here, as I'm not sure how to simplify this equation.
I would really appreciate if you help me on this. Thanks.
• Mar 24th 2008, 07:20 PM
mr fantastic
Quote:

Originally Posted by rednest
Since
$y = - \frac{b \cos t}{a \sin t} (x - a \cos t) + b \sin t$
$y = \frac{b \sin t}{a \cos t} (x + a \sin t) + b \cos t$
I let $\frac{b \sin t}{a \cos t} (x + a \sin t) + b \cos t = - \frac{b \cos t}{a \sin t} (x - a \cos t) + b \sin t$.

But I'm stuck here, as I'm not sure how to simplify this equation.
I would really appreciate if you help me on this. Thanks.

Have you tried expanding, collecting like terms, simplifying .....?
• Mar 24th 2008, 09:31 PM
rednest
Quote:

Originally Posted by mr fantastic
Have you tried expanding, collecting like terms, simplifying .....?

I get $\frac{bxsint}{acost} + \frac{absin^2t}{acost} + \frac{abcos^2t}{acost}$
= $-\frac{bxcost}{asint} + \frac{abcos^2t}{asint} + \frac{absin^2t}{asint}$.

Then I simplify to get $\frac{bxsint}{acost} = -\frac{bxcost}{asint}$, giving me $abx = 0$, which makes no sense to me.
• Mar 25th 2008, 12:27 AM
mr fantastic
Quote:

Originally Posted by rednest
[snip]
$\frac{b \sin t}{a \cos t} (x + a \sin t) + b \cos t = - \frac{b \cos t}{a \sin t} (x - a \cos t) + b \sin t$.

But I'm stuck here, as I'm not sure how to simplify this equation.[snip]

Multiply both sides by $a \cos t \sin t$ to get rid of the fractions:

$b \sin^2 t (x + a \sin t) + ab \cos^2 t \sin t = -b \cos^2 t (x - a \cos t) + ab \sin^2 t \cos t$.

After expanding both sides, grouping x-terms and simplifying using the identity $\sin^2 t + \cos^2 t = 1$ you get

$x = a (\cos^3 t - \cos^2 t \sin t + \sin^2 t \cos t - \sin^3 t)$

$= a (\cos^2 t + \sin^2 t) (\cos t - \sin t) = a(\cos t - \sin t)$.

Substitute into either of equations (1) or (2) (see my first reply) to get $y = b(\cos t + \sin t)$.

So you have:

$\frac{x}{a} = \cos t - \sin t$ .... (3)

$\frac{y}{b} = \cos t + \sin t$ .... (4)

You should be able to easily eliminate t from equations (3) and (4) to get $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 2$.