What is the vertices 7x^2+16y^2=112 ?

The vertices 7x^2+16y^2=112 is (4,0),(-4,0)

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vertices 7x^2+16y^2=112

vertices

7 x^{2} + 16 y^{2} = 112

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

Solution steps

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

Calculate ellipse properties

7 x^{2} + 16 y^{2} = 112 :

Ellipse with center

(h, k) = (0, 0), a = 4, b = 7

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

Refine

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

9 x^{2} + 49 y^{2} = 441

16 x^{2} + 9 y^{2} - 96 x + 36 y - 36 = 0

\frac{x ^{2}}{7 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

25 x^{2} + 9 y^{2} - 250 x + 72 y + 544 = 0

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

What is the vertices 7x^2+16y^2=112 ?
The vertices 7x^2+16y^2=112 is (4,0),(-4,0)