equation of the ellipse

[Guest]

use the definition of an ellipse to find an equation of the ellipse having foci (1,1) and (-1,-1) and sum of focal radii 3.

the answer to this problem is : 20x^2 + 20y^2-32xy-9=0

I tried attempting the problem. Here's the work:

2a=3
a=3/2

using the distance formula and dividing this by 2, i found c= sqrt (2)

then i plugged a and c into the formula, b^2=a^2-c^2 and b^2=9/4 - sqrt (2)

 

[soroban]

Hello, bob1!

This requires the Distance Formula and a lot of careful algebra . . .

Use the definition of an ellipse to find an equation of the ellipse

having foci $\displaystyle F_1(1,\,1)$ and $\displaystyle F_2(-1,\,-1)$ and sum of focal radii 3.

The answer to this problem is: $\displaystyle 20x^2\,+\,20y^2\,-\,32xy\,-\,9\:=\:0\;$

Let $\displaystyle P(x,y)$ be a point on the ellipse . . . then: $\displaystyle \,\overline{PF}_1\,+\,\overline{PF}_2\:=\:3$

Then we have: \(\displaystyle \,\sqrt{(x\,-\,1)^2\,+\,(y\,-\,1)^2}\.+\,\sqrt{(x\,+\,1)^2\,+\,(y\,+\,1)^2}\;=\;3\)

Isolate a radical: $\displaystyle \,\sqrt{(x\,-\,1)^2\,+\,(y\,-\,1)^2}\;=\;3\,-\,\sqrt{(x\,+\,1)^2\;+\;(y\,+\,1)^2}$

Square both sides: $\displaystyle \,(x\,-\,1)^2\,+\,(y\,-\,1)^2\;=\;9\,-\,6\sqrt{(x\,+\,1)^2\,+\,(y\,+\,1)^2}\,+\,(x\,+\,1)^2\,+\,(y\,+\,1)^2$

Expand: $\displaystyle \,\not{x^2}\,-\,2x\,+\,\not{1}\,+\,\not{y^2}\,-\,2y\,+\,\not{1}\;=\;9\,-\,6\sqrt{(x\,+\,1)^2\,+\,(y\,+\,1)^2}\,+\,\not{x^2}\,+\,2x\,+\,\not{1}\,+\,\not{y^2}\,+\,2y\,+\,\not{1}$

Simplify: $\displaystyle \,6\sqrt{(x\,+\,1)^2\,+\,(y\,+\,1)^2} \;= \;4(x\,+\,y)\,+\,9$

Square both sides: $\displaystyle \,36\left[(x\,+\,1)^2\,+\,(y\,+\,1)^2\right]\;=\;16(x\,+\,y)^2\,+\,72(x\,+\,y)\,+\,81$

Expand: $\displaystyle \,36x^2\,+\,\sout{72x}\,+\,36\,+\,36y^2\,+\,\sout{72y}\,+\,36\;=\;16x^2\,+\,32xy\,+\,16y^2\,+\,\sout{72x}\,+\,\sout{72y}\,+\,81$

Therefore: $\displaystyle \:20x^2\,+\,20y^2\,-\,32xy\,-\,9\;=\;0\;\;$ . . . ta-DAA!