Question

Find the equation for the ellipse that satisfies the given conditions: Length of minor axis $16$, foci $(0,\pm 6)$.

Answer 1

Given: Conditions for ellipse:
Length of minor axis $16$, foci $(0,\pm 6)$.
$\because$ Foci are on $y$-axis.
$\therefore$ The major axis is along the $y$-axis.
Hence, equation of ellipse is of the form: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Given that,
Length of the minor axis is $16$.
i.e., $2b=16\Rightarrow b=8$
Foci $=(0,\pm c)=(0,\pm 6)$ (given)
$\therefore c=6$
$b=8,c=6$
We know, $c^2=a^2-b^2$
$\Rightarrow 6^2=a^2-8^2$
$\Rightarrow 36=a^2-64$
$\Rightarrow a^2=64+36$
$\Rightarrow a^2=100$
$\Rightarrow a=\sqrt{100}=10$
Putting value of $a$ and $b$ in $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
$\therefore$ Required equation of the ellipse is $\frac{x^2}{8^2}+\frac{y^2}{{10}^2}=1$ or $\frac{x^2}{64}+\frac{y^2}{100}=1$