Question

Find the equation of ellipse whose length of minor axis is $16$ and foci is $(0,\pm 6)$

Answer 1

Length of minor axis $=16$

and foci $=(0,\pm 6).......\left(1\right)$

Since foci is of type $(0,\pm c)....\left(2\right)$

The major axis is along the y-axis and required equation of ellipse is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

From (1) and (2)

$c=6$

Length of minor axis $=2b$

$16=2b$

$b=8$

Also, we know

$c^2=a^2-b^2$

$(6{)}^2=a^2-(8{)}^2$

$a^2=36+64$

$a^2=100$

$a=\sqrt{100}$

$a=10$

Equation of ellipse is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

putting values

$\frac{x^2}{8^2}+\frac{y^2}{{10}^2}=1$

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

This is the required equation.