Analytic Geometry: Proving R Lies on Ellipse

[dajugganaut]

Hi all. I have a analytic geometry question that I need a bit of help with.
consider the concentric circles with the equations:

$$x^2 + y^2 = 9$$
and
$$x^2 +y^2 = 4$$

A radius from the center O intersects the inner circle at P and the outer circle at Q. The line parallel to the x-axis through P meets the line parallel to the y-axis through Q at the point R. Prove that R lies on the ellipse
$$((x^2)/9) + ((y^2)/4) =1$$

Some of the facts I've established:

the distance from P to Q is 1 for sure. Also, the triagle PQR is a right angle triangle. using the ellipse equation, i know that the distance from the center to the vertex is 3, and that the distance from the center to the focus is the square root of 5.

 

[calculus1967]

Why wouldn't the distance of PQ be 5?

 

[dajugganaut]

if the radius of the inner circle is 2 and the outer one is 3, is the distance between them not 1?

 

[calculus1967]

Yeah, for some reason I wasn't thinking about those being the squares of the radii. Anyway, I think $R$ would have the ascissa of $Q$ and the ordinate of $P$. If you solve the respective equations for them, you find $(\sqrt{9-y^2}, \sqrt{4-x^2})$. I would try solve the equation of ellipse for x and then y and it should be equivalent to the coordinates. I'm not very good at making proofs though, so there might be a problem with that way.

 

[HallsofIvy]

Write the point P as $$(x_0,y_0)$$ and Q as $$(x_1,y_1)$$. The The fact that P and Q lie on the same radius means that $$\frac{y_0}{x_0}= \frac{y_1}{x_1}$$. The horizontal line through P (parallel to the x-axis) is y= y₀ and the vertical line through Q (parallel to the y-axis) is x= x₁. The point R has coordinates (x₁,y₀). Of course, $$x_0^2+ y_0^2= 4$$ and $$x_1^2+ y_1^2= 9$$. Thats a total of 3 equation is 4 unknowns. Use the equations to eliminate x₁ and y₀. I wouldn't be at all surprized if you were left with $$\frac{x_0^2}{9}+{y_1^2}{4}= 1$$!