What is the vertices ((x-6)^2)/(49)+((y-6)^2)/(16)=1 ?

The vertices ((x-6)^2)/(49)+((y-6)^2)/(16)=1 is (13,6),(-1,6)

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vertices ((x-6)^2)/(49)+((y-6)^2)/(16)=1

vertices

\frac{( x - 6 ) ^{2}}{49} + \frac{( y - 6 ) ^{2}}{16} = 1

(13, 6), (- 1, 6)

Solution steps

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

Calculate ellipse properties

\frac{( x - 6 ) ^{2}}{49} + \frac{( y - 6 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (6, 6), a = 7, b = 4

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

Refine

(13, 6), (- 1, 6)

\frac{( x + 1 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{49} = 1

\frac{( x - 1 ) ^{2}}{6} + \frac{( y + 2 ) ^{2}}{18} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{52} = 1

x^{2} + 9 y^{2} + 16 x + 18 y + 64 = 0

x^{2} + 3 y^{2} + 6 x - 18 y + 33 = 0

What is the vertices ((x-6)^2)/(49)+((y-6)^2)/(16)=1 ?
The vertices ((x-6)^2)/(49)+((y-6)^2)/(16)=1 is (13,6),(-1,6)