What is the vertices 16x^2-12y^2=192 ?

The vertices 16x^2-12y^2=192 is (2sqrt(3),0),(-2sqrt(3),0)

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vertices 16x^2-12y^2=192

vertices

16 x^{2} - 12 y^{2} = 192

(23, 0), (- 23, 0)

Solution steps

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

Calculate Hyperbola properties

16 x^{2} - 12 y^{2} = 192 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 23, b = 4

(0 + 23, 0), (0 - 23, 0)

Refine

(23, 0), (- 23, 0)

x^{2} - y^{2} = 14

\frac{( x - 2 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{16} = 1

\frac{y ^{2}}{4} - \frac{x ^{2}}{25} = 1

\frac{x ^{2}}{7} - \frac{7 y ^{2}}{42} = 1

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

What is the vertices 16x^2-12y^2=192 ?
The vertices 16x^2-12y^2=192 is (2sqrt(3),0),(-2sqrt(3),0)