What is the vertices 7x^2+16y^2-112=0 ?

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

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vertices 7x^2+16y^2-112=0

vertices

7 x^{2} + 16 y^{2} - 112 = 0

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

Solution steps

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

Calculate ellipse properties

7 x^{2} + 16 y^{2} - 112 = 0 :

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} + 25 y^{2} + 90 x = 0

2 x^{2} + 3 y^{2} + 4 x - 6 y - 1 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1

36 x^{2} + 9 y^{2} = 324

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

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