I need help with an ellipse question. URGENT!

[Guest]

Ok here's the question:

An ellipse has foci f1(3,0) and f2(3,0). Determine the equation of the ellipse so that quadrilateral f1b1f2b2 is a square.

I URGENTLY need this tonight!

 

[Unco]

You have \(\displaystyle \mbox{a}\). Draw the square (side length = ?), it can go above or below the x-axis. Then what is \(\displaystyle \mbox{b}\)?

I am a little confused by the order \(\displaystyle \mbox{f_1b_1f_2b_2 }\); sure it's not \(\displaystyle \mbox{f_1b_1b_2f_2}\) or similar?

 

[soroban]

Hello, breanne!

I always ask: Did you make a sketch?

You gave the same focus twice . . . I assume it's a typo.
I also assume that \(\displaystyle b_1,\,b_2\) are the \(\displaystyle b\)-values in: \(\displaystyle \L\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\:=\:1\)

An ellipse has foci \(\displaystyle f_1(3,\,0)\) and \(\displaystyle f_2(-3,\,0)\)

Determine the equation of the ellipse so that quadrilateral \(\displaystyle f_1b_1f_2b_2\) is a square.

                b1
              * * *
         *    / | \    *
     *      /   |   \      *
    *     /     |     \     *
 a2 *---*-------+-------*---* a1
    *  f2 \     |     / f1  *
     *      \   |   /      *
         *    \ | /    *
              * * *
                b2

Once you make a sketch, the answer is almost obvious . . .

They gave us: \(\displaystyle \,c\,=\,3\)

If that quadrilateral is a square, then \(\displaystyle \,b\,=\,3\)

From \(\displaystyle \,a^2\:=\:b^2\,+\,c^2\), we have: \(\displaystyle \,a^2\:=\:3^2\,+\,3^2\:=\:18\)

The equation of the ellipse is: \(\displaystyle \L\,\frac{x^2}{18}\,+\,\frac{y^2}{9}\:=\:1\)

 

[Guest]

oh.. I'm so stupid. Thanks so much