conic sections

• Mar 2nd 2008, 03:41 PM
brittneyv
conic sections
find the equation of the ellipse that goes through the point (2,3) and has foci (2,0) and (-2,0).
• Mar 2nd 2008, 06:14 PM
Soroban
Hello, brittneyv!

Quote:

Find the equation of the ellipse that goes through the point $(2,3)$
and has foci $(\pm2,0)$

The "focus equation" is: . $a^2 \:=\:b^2+c^2$
. . Since $c = \pm2$, we have: . $a^2 \:=\:b^2+4\;\;{\color{blue}[1]}$

The ellipse has the form: . $\frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1\quad\Rightarrow\quad b^2x^2 + a^2y^2 \:=\:a^2b^2$

Since $(2,3)$ is on the ellipse, we have: . $4b^2 + 9a^2 \:=\:a^2b^2\;\;{\color{blue}[2]}$

Substitute [1] into [2]: . $4b^2 + 9(b^2+4) \:=\:b^2(b^2+4)$
. . which simplifies to: . $b^4 - 9b^2 - 36 \:=\:0\quad\Rightarrow\quad (b^2-12)(b^2+ 3)\:=\:0$
Hence: . $b^2 \:=\:12$

Substitute into [1]: . $a^2 \:=\:12 + 4 \:=\:16$

Therefore, the equation is: . $\boxed{\frac{x^2}{16} + \frac{y^2}{12} \;=\;1}$

• Mar 2nd 2008, 08:15 PM
brittneyv
wow thank you so much!