# Math Help - Coordinate system help

1. ## 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 $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 2$.

2. 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.

3. ## 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.

4. 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.

5. 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.

6. 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 .....?

7. 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.

8. 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 $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 2$.