
Another ellipse question
Stuck on my homework question...again lol...
The tangents of the ellipse with equation x^2/(a^2)+y^2/(b^2)=1 at the points P(acost, bsint) and Q(asint, bcost) intersect at the point R. (I tried to find R by calculating the equation of the two tangents and then equating them to obtain the x and y coordinates). As t varies, show that R lies on the curve with equation x^2/(a^2)+y^2/(b^2)=2. I then tried to eliminate t to get an equation with just x's and y's but I couldn't eliminate t, so I must have gone wrong. When I tried to get the xcoordinate of R, everything cancelled to 0.
Could someone help please?

Why try to eliminate the parameter? There is a reason why folks got into parametric equations. Sometimes, you can do more things.
Find the general slope.
$\displaystyle \frac{dy}{dx}\;=\;\frac{b^{2}x}{a^{2}y}$
You have to know where that came from.
Now, write the equations of the tangent lines.
$\displaystyle yb\sin(t)\;=\;\frac{dy}{dx}(xa\cos(t))$
$\displaystyle yb\cos(t)\;=\;\frac{dy}{dx}(x+a\sin(t))$
Don't forget to substitute the right values in dy/dx.
Solve system, for x and y.
I get x = a(cos(t)sin(t)) and y = b(cos(t)+sin(t))
One last thing. Is that point on the new ellipse?
This is a great review problem. There is a ton of mathematics in there. Understand this problem and you will have some good information for the future.

Hello, eveyone!
When working with tangents to conic curves,
. . there are some formulas that can save a lot of time and work.
Given the point $\displaystyle P(x_1,y_1)$ on the circle: .$\displaystyle x^2 + y^2 \:=\:r^2$
. . the equation of the tangent at P is: .$\displaystyle x_1x + y_1y \:=\:r^2$
Given the point $\displaystyle P(x_1,y_1)$ on the ellipse: .$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1$
. . the equation of the tangent at P is: .$\displaystyle \frac{x_1x}{a^2} + \frac{y_1y}{b^2} \:=\:1$
Given the point $\displaystyle P(x_1,y_1)$ on the hyperbola: .$\displaystyle \frac{x^2}{a^2}  \frac{y^2}{b^2} \:=\:1$
. . the equation of the tangent at P is: .$\displaystyle \frac{x_1x}{a^2}  \frac{y_1y}{b^2} \:=\:1$