Question

determine the foci of an ellipse
find the foci of the ellipse with the equation $\frac{x^2}{9} + \frac{y^2}{36} = 1$.
in this example, you are not given the values for $a$ and $b$, but there is enough information to find them.
look at the denominators in this equation. the greater number is 36, and it is under $y^2$. therefore, the major axis is on the $y$-axis, so it is vertical.
if $a^2 = 36$, then $a = 6$. therefore, the vertices are $(0, 6)$ and $(0, -6)$.
if $b^2 = 9$, then $b = 3$. therefore, the co - vertices are $(3, 0)$ and $(-3, 0)$.
here is the graph of the ellipse.
next, you will determine the foci for this ellipse.
type your answer and then click or tap done.
using the equation $c^2 = a^2 - b^2$ find the value of $c$. simplify the radical if necessary.
$c = \square$
ellipses

Explanation:

Step1: Identificar $a^2$ y $b^2$

De la ecuación $\frac{x^2}{9} + \frac{y^2}{36} = 1$, se tiene $a^2=36$ y $b^2=9$.

Step2: Calcular $c$ usando la fórmula

Usar $c^2 = a^2 - b^2$:
$c^2 = 36 - 9 = 27$
$c = \sqrt{27} = 3\sqrt{3}$

Step3: Determinar los focos

Como el eje mayor es vertical, los focos están en $(0, c)$ y $(0, -c)$, es decir $(0, 3\sqrt{3})$ y $(0, -3\sqrt{3})$.

Answer:

Los focos son $(0, 3\sqrt{3})$ y $(0, -3\sqrt{3})$, y $c=3\sqrt{3}$