What is the vertices 16x^2+25y^2-1600=0 ?

The vertices 16x^2+25y^2-1600=0 is (10,0),(-10,0)

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vertices 16x^2+25y^2-1600=0

vertices

16 x^{2} + 25 y^{2} - 1600 = 0

(10, 0), (- 10, 0)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} - 1600 = 0 :

Ellipse with center

(h, k) = (0, 0), a = 10, b = 8

(0 + 10, 0), (0 - 10, 0)

Refine

(10, 0), (- 10, 0)

\frac{x ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

x^{2} + 10 x + 9 y^{2} + 16 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

4 x^{2} + 9 y^{2} - 8 x - 36 y + 4 = 0

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

What is the vertices 16x^2+25y^2-1600=0 ?
The vertices 16x^2+25y^2-1600=0 is (10,0),(-10,0)