What is the vertices (x/7)^2+(y/5)^2=1 ?

The vertices (x/7)^2+(y/5)^2=1 is (7,0),(-7,0)

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vertices (x/7)^2+(y/5)^2=1

vertices

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

(7, 0), (- 7, 0)

Solution steps

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

Calculate ellipse properties

(\frac{x}{7})^{2} + (\frac{y}{5})^{2} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 7, b = 5

(0 + 7, 0), (0 - 7, 0)

Refine

(7, 0), (- 7, 0)

36 x^{2} + 25 y^{2} - 50 y - 875 = 0

4 (x - 5)^{2} + 3 (y - 5)^{2} = 192

9 x^{2} + 16 y^{2} + 36 x - 160 y + 292 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{81} = 1

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

What is the vertices (x/7)^2+(y/5)^2=1 ?
The vertices (x/7)^2+(y/5)^2=1 is (7,0),(-7,0)