What is the vertices 4x^2-3y^2-8x-8=0 ?

The vertices 4x^2-3y^2-8x-8=0 is (1+sqrt(3),0),(1-sqrt(3),0)

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vertices 4x^2-3y^2-8x-8=0

vertices

4 x^{2} - 3 y^{2} - 8 x - 8 = 0

(1 + 3, 0), (1 - 3, 0)

Solution steps

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

Calculate Hyperbola properties

4 x^{2} - 3 y^{2} - 8 x - 8 = 0 :

Right-left Hyperbola with

(h, k) = (1, 0), a = 3, b = 2

(1 + 3, 0), (1 - 3, 0)

\frac{x ^{2}}{2} - \frac{y ^{2}}{3} = 1

- 4 x^{2} + y^{2} + 16 x + 12 y + 16 = 0

\frac{y ^{2}}{81} - \frac{x ^{2}}{64} = 1

\frac{( x ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{9} = 1

4 (y - 3)^{2} - 9 (x - 2)^{2} = 36

What is the vertices 4x^2-3y^2-8x-8=0 ?
The vertices 4x^2-3y^2-8x-8=0 is (1+sqrt(3),0),(1-sqrt(3),0)