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

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

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

vertices

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

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

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} > 1

12 x^{2} + 20 y^{2} - 12 x + 40 y - 37 = 0

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

(\frac{x ^{2}}{15}) + (\frac{y ^{2}}{16}) = 1

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

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