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

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

(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)

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400

3 x^{2} + 4 y^{2} - 12 x = 36

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

\frac{x ^{2}}{9} + \frac{( y + 5 ) ^{2}}{25} = 1

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)