What is the vertices x^2+8y^2=32 ?

The vertices x^2+8y^2=32 is (4sqrt(2),0),(-4sqrt(2),0)

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vertices x^2+8y^2=32

vertices

x^{2} + 8 y^{2} = 32

(42, 0), (- 42, 0)

Solution steps

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

Calculate ellipse properties

x^{2} + 8 y^{2} = 32 :

Ellipse with center

(h, k) = (0, 0), a = 42, b = 2

(0 + 42, 0), (0 - 42, 0)

Refine

(42, 0), (- 42, 0)

2 x^{2} + y^{2} + 12 x - 6 y + 26 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{4} = 1

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

25 x^{2} + 4 y^{2} = 100

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

What is the vertices x^2+8y^2=32 ?
The vertices x^2+8y^2=32 is (4sqrt(2),0),(-4sqrt(2),0)