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$\frac{x^2}{64}+\frac{y^2}{100}=1$