What is the vertices ((x-3)^2)/(25)+(y-2)^2=1 ?

The vertices ((x-3)^2)/(25)+(y-2)^2=1 is (8,2),(-2,2)

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vertices ((x-3)^2)/(25)+(y-2)^2=1

vertices

\frac{( x - 3 ) ^{2}}{25} + (y - 2)^{2} = 1

(8, 2), (- 2, 2)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{25} + (y - 2)^{2} = 1 :

Ellipse with center

(h, k) = (3, 2), a = 5, b = 1

(3 + 5, 2), (3 - 5, 2)

Refine

(8, 2), (- 2, 2)

x^{2} + 4 y^{2} + 2 x - 12 y + 6 = 0

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{5 x ^{2} + 8 y ^{2} - 1}{11 x ^{2} - 13 y ^{2}} = 0

\frac{y ^{2}}{16} + \frac{x ^{2}}{49} = 1

What is the vertices ((x-3)^2)/(25)+(y-2)^2=1 ?
The vertices ((x-3)^2)/(25)+(y-2)^2=1 is (8,2),(-2,2)