What is the vertices ((x+6)^2)/(36)+((y-7)^2)/(32)=1 ?

The vertices ((x+6)^2)/(36)+((y-7)^2)/(32)=1 is (0,7),(-12,7)

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vertices ((x+6)^2)/(36)+((y-7)^2)/(32)=1

vertices

\frac{( x + 6 ) ^{2}}{36} + \frac{( y - 7 ) ^{2}}{32} = 1

(0, 7), (- 12, 7)

Solution steps

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

Calculate ellipse properties

\frac{( x + 6 ) ^{2}}{36} + \frac{( y - 7 ) ^{2}}{32} = 1 :

Ellipse with center

(h, k) = (- 6, 7), a = 6, b = 42

(- 6 + 6, 7), (- 6 - 6, 7)

Refine

(0, 7), (- 12, 7)

16 x^{2} + 4 y^{2} + 96 x - 24 y + 116 = 0

\frac{( x - 4 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{64} = 1

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

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

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

What is the vertices ((x+6)^2)/(36)+((y-7)^2)/(32)=1 ?
The vertices ((x+6)^2)/(36)+((y-7)^2)/(32)=1 is (0,7),(-12,7)