What is the vertices x^2= 9/16 y^2+9 ?

The vertices x^2= 9/16 y^2+9 is (3,0),(-3,0)

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vertices x^2= 9/16 y^2+9

vertices

x^{2} = \frac{9}{16} y^{2} + 9

(3, 0), (- 3, 0)

Solution steps

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

Calculate Hyperbola properties

x^{2} = \frac{9}{16} y^{2} + 9 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 4

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

9 x^{2} - 4 y^{2} - 36 x + 8 y - 4 = 0

\frac{x ^{2}}{9} - \frac{y ^{2}}{1} = 1

y^{2} - x^{2} = 1

\frac{( x - 5 ) ^{2}}{81} - \frac{( y - 7 ) ^{2}}{49} = 1

\frac{x ^{2}}{25} - \frac{y ^{2}}{64} = 1

What is the vertices x^2= 9/16 y^2+9 ?
The vertices x^2= 9/16 y^2+9 is (3,0),(-3,0)