What is the vertices (x^2)/(49)+((y-3)^2)/(16)=1 ?

The vertices (x^2)/(49)+((y-3)^2)/(16)=1 is (7,3),(-7,3)

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vertices (x^2)/(49)+((y-3)^2)/(16)=1

vertices

(7, 3), (- 7, 3)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{49} + \frac{( y - 3 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 3), a = 7, b = 4

(0 + 7, 3), (0 - 7, 3)

Refine

(7, 3), (- 7, 3)

What is the vertices (x^2)/(49)+((y-3)^2)/(16)=1 ?
The vertices (x^2)/(49)+((y-3)^2)/(16)=1 is (7,3),(-7,3)