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

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

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

vertices

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

(6 + 2, 2), (6 - 2, 2)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (6, 2), a = 2, b = 1

(6 + 2, 2), (6 - 2, 2)

4 x^{2} + y^{2} = 4 x

4 x^{2} + 4 x + y^{2} = 0

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

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{10} = 1

\frac{x ^{2}}{9} + \frac{( y - 4 ) ^{2}}{7} = 1

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