What is the vertices (x^2)/(25)+(y^2)/(16)=1 ?

The vertices (x^2)/(25)+(y^2)/(16)=1 is (5,0),(-5,0)

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vertices (x^2)/(25)+(y^2)/(16)=1

vertices

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 5, b = 4

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

Refine

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

4 \leq - x^{2} - y

18 x^{2} - 64 x - 14 y + 150 = 0

x^{2} + y^{2} + 6 x - 4 y - 12 = 0

36 x^{2} + y^{2} = 36

x^{2} - 4 x + y^{2} + 6 y - 12 = 0

What is the vertices (x^2)/(25)+(y^2)/(16)=1 ?
The vertices (x^2)/(25)+(y^2)/(16)=1 is (5,0),(-5,0)