Question

1. find the foci of the ellipse with the equation \\(\frac{x^2}{9} + \frac{y^2}{25} = 1\\)

(0, +4)
(±4, 0)
(±2√3, 0)
(0, ±3√5)

Explanation:

Step1: Identify ellipse parameters

The standard form of a vertical major axis ellipse is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$. For $\frac{x^2}{9}+\frac{y^2}{25}=1$, we have $a^2=25$, $b^2=9$, so $a=5$, $b=3$.

Step2: Calculate focal distance $c$

Use the ellipse relationship $c^2=a^2-b^2$.
$c^2=25-9=16$, so $c=4$.

Step3: Locate the foci

Since the major axis is vertical, foci are at $(0, \pm c)=(0, \pm4)$.

Answer:

(0, ±4)