What is the vertices x^2+5x-y^2+3y=1 ?

The vertices x^2+5x-y^2+3y=1 is ((-5+2sqrt(5))/2 , 3/2),((-5-2sqrt(5))/2 , 3/2)

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vertices x^2+5x-y^2+3y=1

vertices

x^{2} + 5 x - y^{2} + 3 y = 1

(\frac{- 5 + 2 5}{2}, \frac{3}{2}), (\frac{- 5 - 2 5}{2}, \frac{3}{2})

Solution steps

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

Calculate Hyperbola properties

x^{2} + 5 x - y^{2} + 3 y = 1 :

Right-left Hyperbola with

(h, k) = (- \frac{5}{2}, \frac{3}{2}), a = 5, b = 5

(- \frac{5}{2} + 5, \frac{3}{2}), (- \frac{5}{2} - 5, \frac{3}{2})

Refine

(\frac{- 5 + 2 5}{2}, \frac{3}{2}), (\frac{- 5 - 2 5}{2}, \frac{3}{2})

9 x^{2} - 4 y^{2} + 36 x + 8 y + 33 = 0

- 4 x^{2} + 24 x + 16 y^{2} - 128 y + 156 = 0

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

\frac{( x - 3 ) ^{2}}{49} - \frac{( y - 2 ) ^{2}}{16} = 1

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

What is the vertices x^2+5x-y^2+3y=1 ?
The vertices x^2+5x-y^2+3y=1 is ((-5+2sqrt(5))/2 , 3/2),((-5-2sqrt(5))/2 , 3/2)