What is the vertices 15x^2+16y^2=240 ?

The vertices 15x^2+16y^2=240 is (4,0),(-4,0)

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vertices 15x^2+16y^2=240

vertices

15 x^{2} + 16 y^{2} = 240

(4, 0), (- 4, 0)

Solution steps

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

Calculate ellipse properties

15 x^{2} + 16 y^{2} = 240 :

Ellipse with center

(h, k) = (0, 0), a = 4, b = 15

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

\frac{x ^{2}}{49} + \frac{( y - 3 ) ^{2}}{16} = 1

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400

3 x^{2} + 4 y^{2} - 12 x = 36

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

What is the vertices 15x^2+16y^2=240 ?
The vertices 15x^2+16y^2=240 is (4,0),(-4,0)