What is the vertices (x^2}{25}+\frac{y^2)/9 =1 ?

The vertices (x^2}{25}+\frac{y^2)/9 =1 is (5,0),(-5,0)

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vertices (x^2}{25}+\frac{y^2)/9 =1

vertices

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 5, b = 3

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

Refine

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

x^{2} - \frac{y ^{2}}{16} = 1

f (x) = (x - 5) (x + 1)

x^{2} + (y - 4)^{2} = 4

x^{2} - 6 x + 5 y = - 34

f (x) = x^{2} - 2 x - 3

What is the vertices (x^2}{25}+\frac{y^2)/9 =1 ?
The vertices (x^2}{25}+\frac{y^2)/9 =1 is (5,0),(-5,0)