Question

The coordinates of the foci of the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$ is/are

• A. $(\sqrt{5},0)$
• B. $(2\sqrt{5},0)$
• C. $(-\sqrt{5},0)$
• D. $(-2\sqrt{5},0)$

Answer 1

Answer: (B), (D)

$(-2\sqrt{5},0)$

Given parabola
$\frac{x^2}{36}+\frac{y^2}{16}=1$
Which is in the form of
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a=6,b=4$
$e^2=1-\frac{b^2}{a^2}\Rightarrow =1-\frac{16}{36}\Rightarrow e=\frac{\sqrt{5}}{3}$
Foci $=(\pm ae,0)=(\pm 2\sqrt{5},0)$