# Another ellipse question

• Dec 19th 2007, 11:13 AM
free_to_fly
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 x-coordinate of R, everything cancelled to 0.

• Dec 19th 2007, 01:04 PM
TKHunny
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.

$\frac{dy}{dx}\;=\;-\frac{b^{2}x}{a^{2}y}$

You have to know where that came from.

Now, write the equations of the tangent lines.

$y-b\sin(t)\;=\;\frac{dy}{dx}(x-a\cos(t))$

$y-b\cos(t)\;=\;\frac{dy}{dx}(x+a\sin(t))$

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.
• Dec 20th 2007, 01:32 PM
Soroban
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 $P(x_1,y_1)$ on the circle: . $x^2 + y^2 \:=\:r^2$

. . the equation of the tangent at P is: . $x_1x + y_1y \:=\:r^2$

Given the point $P(x_1,y_1)$ on the ellipse: . $\frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1$

. . the equation of the tangent at P is: . $\frac{x_1x}{a^2} + \frac{y_1y}{b^2} \:=\:1$

Given the point $P(x_1,y_1)$ on the hyperbola: . $\frac{x^2}{a^2} - \frac{y^2}{b^2} \:=\:1$

. . the equation of the tangent at P is: . $\frac{x_1x}{a^2} - \frac{y_1y}{b^2} \:=\:1$