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

The vertices (y^2)/(49)-(x^2)/(16)=1 is (0,7),(0,-7)

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

vertices

\frac{y ^{2}}{49} - \frac{x ^{2}}{16} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{49} - \frac{x ^{2}}{16} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 7, b = 4

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

Refine

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

- 3 x^{2} + 2 y^{2} + 6 x - 7 y = 7

\frac{( y - 3 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{16} = 1

x^{2} - y^{2} - 4 x + 8 y - 21 = 0

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

10 x^{2} - 25 y^{2} = 100

What is the vertices (y^2)/(49)-(x^2)/(16)=1 ?
The vertices (y^2)/(49)-(x^2)/(16)=1 is (0,7),(0,-7)