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

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

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

vertices

\frac{x ^{2}}{6 ^{2}} + \frac{y ^{2}}{7 ^{2}} = 1

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

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

\frac{x ^{2}}{6 ^{2}} + \frac{y ^{2}}{7 ^{2}} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 7, a = 6

(0, 0 + 7), (0, 0 - 7)

Refine

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

\frac{x ^{2}}{49} + \frac{y ^{2}}{4} = 1

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

(x^{2} + 8 (y + 5)^{2}) = 20

x^{2} + 4 y^{2} = 4 x + 8

\frac{( x - 1 ) ^{2}}{3} + \frac{( y - 2 ) ^{2}}{5} = 1

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