What is the vertices (x^2}{25}+\frac{y^2)/4 =1 ?

The vertices (x^2}{25}+\frac{y^2)/4 =1 is (5,0),(-5,0)

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vertices (x^2}{25}+\frac{y^2)/4 =1

vertices

\frac{x ^{2}}{25} + \frac{y ^{2}}{4} = 1

(5, 0), (- 5, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

\frac{x ^{2}}{25} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 5, b = 2

(0 + 5, 0), (0 - 5, 0)

Refine

(5, 0), (- 5, 0)

4 x^{2} + y^{2} = 16

y^{2} = - 8 x

x = y^{2} - 5

y = x^{2} + 2 x - 3

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

What is the vertices (x^2}{25}+\frac{y^2)/4 =1 ?
The vertices (x^2}{25}+\frac{y^2)/4 =1 is (5,0),(-5,0)