What is the vertices 4x^2-9y^2-36<0 ?

The vertices 4x^2-9y^2-36<0 is (3,0),(-3,0)

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vertices 4x^2-9y^2-36<0

vertices

4 x^{2} - 9 y^{2} - 36 < 0

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

Solution steps

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

Rewrite as

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 2

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

Refine

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

\frac{x ^{2}}{16} - \frac{y ^{2}}{64} = 1

x^{2} - 3 y^{2} + 2 x + 6 y - 1 = 0

9 x^{2} - 16 y^{2} - 18 x - 64 y - 199 = 0

x^{2} - 9 y^{2} = 9

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

What is the vertices 4x^2-9y^2-36<0 ?
The vertices 4x^2-9y^2-36<0 is (3,0),(-3,0)