Question:

The tangents to the ellipse with equation at the points and intersect at the point . As varies, show that lies on the curve with equation .

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- Mar 24th 2008, 06:23 AMrednestCoordinate system help
Question:

The tangents to the ellipse with equation at the points and intersect at the point . As varies, show that lies on the curve with equation . - Mar 24th 2008, 06:41 AMmr fantastic
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 !

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 .

I'd use implicit differentiation to get .

At the point , .

So the tangent at this point has equation

.... (1)

At the point , .

So the tangent at this point has equation

.... (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, 07:57 AMrednestNext 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, 07:41 PMmr fantastic
- Mar 24th 2008, 08:05 PMrednest
- Mar 24th 2008, 08:20 PMmr fantastic
- Mar 24th 2008, 10:31 PMrednest
- Mar 25th 2008, 01:27 AMmr fantastic
Multiply both sides by to get rid of the fractions:

.

After expanding both sides, grouping x-terms and simplifying using the identity you get

.

Substitute into either of equations (1) or (2) (see my first reply) to get .

So you have:

.... (3)

.... (4)

You should be able to easily eliminate t from equations (3) and (4) to get .