What is the vertices ((x+3)^2)/(26)+((y-4)^2)/(16)=1 ?

The vertices ((x+3)^2)/(26)+((y-4)^2)/(16)=1 is (-3+sqrt(26),4),(-3-sqrt(26),4)

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vertices ((x+3)^2)/(26)+((y-4)^2)/(16)=1

vertices

\frac{( x + 3 ) ^{2}}{26} + \frac{( y - 4 ) ^{2}}{16} = 1

(- 3 + 26, 4), (- 3 - 26, 4)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{26} + \frac{( y - 4 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (- 3, 4), a = 26, b = 4

(- 3 + 26, 4), (- 3 - 26, 4)

\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1

4 x^{2} + y^{2} - 32 x + 16 y + 124 = 0

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

x^{2} + 4 y^{2} - 6 x + 20 y - 2 = 0

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

What is the vertices ((x+3)^2)/(26)+((y-4)^2)/(16)=1 ?
The vertices ((x+3)^2)/(26)+((y-4)^2)/(16)=1 is (-3+sqrt(26),4),(-3-sqrt(26),4)