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 .
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.
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 .