What is the vertices 16x^2+25y^2-400=0 ?

The vertices 16x^2+25y^2-400=0 is (5,0),(-5,0)

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vertices 16x^2+25y^2-400=0

vertices

16 x^{2} + 25 y^{2} - 400 = 0

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

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} - 400 = 0 :

Ellipse with center

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

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

Refine

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

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

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

25 x^{2} + 9 y^{2} - 200 x - 36 y + 211 = 0

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

\frac{x ^{2}}{100} + \frac{y ^{2}}{49} = 1

What is the vertices 16x^2+25y^2-400=0 ?
The vertices 16x^2+25y^2-400=0 is (5,0),(-5,0)