What is the vertices 4y^2-x^2+40y-4x+60=0 ?

The vertices 4y^2-x^2+40y-4x+60=0 is (-2,-2),(-2,-8)

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vertices 4y^2-x^2+40y-4x+60=0

vertices

4 y^{2} - x^{2} + 40 y - 4 x + 60 = 0

(- 2, - 2), (- 2, - 8)

Solution steps

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

Calculate Hyperbola properties

4 y^{2} - x^{2} + 40 y - 4 x + 60 = 0 :

Up-down Hyperbola with

(h, k) = (- 2, - 5), a = 3, b = 6

(- 2, - 5 + 3), (- 2, - 5 - 3)

Refine

(- 2, - 2), (- 2, - 8)

\frac{( x - 3 ) ^{2}}{25} - (\frac{y + 1}{25})^{2} = 1

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 4 ) ^{2}}{25} = 1

10 y^{2} - 5 x^{2} = 50

\frac{x ^{2}}{9} - \frac{y ^{2}}{36} = 1

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

What is the vertices 4y^2-x^2+40y-4x+60=0 ?
The vertices 4y^2-x^2+40y-4x+60=0 is (-2,-2),(-2,-8)