# Find the equation of an ellipse that satisfies the following conditions: Length of minor axis $= 16$ and Foci $= (0,\pm6)$

$\begin{array}{1 1}\frac{x^2}{64}+\frac{y^2}{100}=1 \\ \frac{x^2}{16}+\frac{y^2}{36}=1 \\ \frac{x^2}{64}+\frac{y^2}{36}=1 \\ \frac{x^2}{100}+\frac{y^2}{64}=1\end{array}$

Toolbox:
• Given an ellipse as follows:
• http://clay6.com/mpaimg/Toolbar_8.png
• The equation of the ellipse is $\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
• Compare the given equation to the general equation of the ellipse to infer $a$ and $b$.
• $c = \sqrt {a^2 - b^2}$
• Given minor axis a, and foci c, we can calculate b and substitute back in the equation above.
Length of minor axis $= 16$ and Foci $= (0,\pm6) \rightarrow 2b = 16 \rightarrow b = 8$ and $c = 6$
$c^2 = a^2 + b^2 \rightarrow 36 = a^2 + 64 \rightarrow a = \sqrt 100 = 10$
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{8^2}$$+\large\frac{y^2}{{10}^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{64}$$+\large\frac{y^2}{100}$$=1$