What is the vertices (x^2)/(7^2)+(y^2)/(3^2)=1 ?

The vertices (x^2)/(7^2)+(y^2)/(3^2)=1 is (7,0),(-7,0)

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vertices (x^2)/(7^2)+(y^2)/(3^2)=1

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = 7, b = 3

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

Refine

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

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

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

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{16} = 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{9} = 1

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

What is the vertices (x^2)/(7^2)+(y^2)/(3^2)=1 ?
The vertices (x^2)/(7^2)+(y^2)/(3^2)=1 is (7,0),(-7,0)