Question

Find the equation of the ellipse whose eccentricity is $\frac{4}{5}$ and axes are along the coordinate axes and foci at $(0,\pm 4)$.

• A. $\frac{x^2}{9}+\frac{y^2}{25}=1$
• B. $\frac{x^2}{4}+\frac{y^2}{16}=1$
• C. $\frac{x^2}{9}+\frac{y^2}{16}=1$
• D. $\frac{x^2}{9}+\frac{y^2}{36}=1$

Answer 1

The correct option is A $\frac{x^2}{9}+\frac{y^2}{25}=1$

Let the required equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

According to the problem, the coordinates of the foci are $(0,\pm 4)$.

We know, coordinates of foci are $(0,\pm be)$.

Therefore, $be=4$

$b\left(\frac{4}{5}\right)=4$

$b=5$

$b^2=25$

Now, $a^2=b^2(1-e^2)$

$a^2=5^2(1-\frac{16}{25})$

$a^2=9$

Thus, the required equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{25}=1$.