What is the vertices x^2-y^2+7x+8y=5 ?

The vertices x^2-y^2+7x+8y=5 is ((-7+sqrt(5))/2 ,4),((-7-sqrt(5))/2 ,4)

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vertices x^2-y^2+7x+8y=5

vertices

x^{2} - y^{2} + 7 x + 8 y = 5

(\frac{- 7 + 5}{2}, 4), (\frac{- 7 - 5}{2}, 4)

Solution steps

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

Calculate Hyperbola properties

x^{2} - y^{2} + 7 x + 8 y = 5 :

Right-left Hyperbola with

(h, k) = (- \frac{7}{2}, 4), a = \frac{5}{2}, b = \frac{5}{2}

(- \frac{7}{2} + \frac{5}{2}, 4), (- \frac{7}{2} - \frac{5}{2}, 4)

Refine

(\frac{- 7 + 5}{2}, 4), (\frac{- 7 - 5}{2}, 4)

16 x^{2} - 25 y^{2} = 400

x^{2} - 9 y^{2} - 4 x + 36 y - 41 = 0

\frac{x ^{2}}{9} - \frac{y ^{2}}{49} = 1

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{49} = 1

What is the vertices x^2-y^2+7x+8y=5 ?
The vertices x^2-y^2+7x+8y=5 is ((-7+sqrt(5))/2 ,4),((-7-sqrt(5))/2 ,4)