What is the vertices 4x^2-y^2+32x+10y+38=0 ?

The vertices 4x^2-y^2+32x+10y+38=0 is (-7/2 ,5),(-9/2 ,5)

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vertices 4x^2-y^2+32x+10y+38=0

vertices

4 x^{2} - y^{2} + 32 x + 10 y + 38 = 0

(- \frac{7}{2}, 5), (- \frac{9}{2}, 5)

Solution steps

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

Calculate Hyperbola properties

4 x^{2} - y^{2} + 32 x + 10 y + 38 = 0 :

Right-left Hyperbola with

(h, k) = (- 4, 5), a = \frac{1}{2}, b = 1

(- 4 + \frac{1}{2}, 5), (- 4 - \frac{1}{2}, 5)

Refine

(- \frac{7}{2}, 5), (- \frac{9}{2}, 5)

4 x^{2} - y^{2} + 12 x + 5 = 0

\frac{y ^{2}}{36} - \frac{x ^{2}}{4} = 1

\frac{( x ^{2} )}{36} - \frac{( y ^{2} )}{25} = 1

16 x^{2} - y^{2} - 64 x - 2 y + 47 = 0

\frac{y ^{2}}{4 ^{2}} - \frac{x ^{2}}{3 ^{2}} = 1

What is the vertices 4x^2-y^2+32x+10y+38=0 ?
The vertices 4x^2-y^2+32x+10y+38=0 is (-7/2 ,5),(-9/2 ,5)