What is the vertices 4x^2-5y^2+30y=65 ?

The vertices 4x^2-5y^2+30y=65 is (sqrt(5),3),(-sqrt(5),3)

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vertices 4x^2-5y^2+30y=65

vertices

4 x^{2} - 5 y^{2} + 30 y = 65

(5, 3), (- 5, 3)

Solution steps

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

Calculate Hyperbola properties

4 x^{2} - 5 y^{2} + 30 y = 65 :

Right-left Hyperbola with

(h, k) = (0, 3), a = 5, b = 2

(0 + 5, 3), (0 - 5, 3)

Refine

(5, 3), (- 5, 3)

f (x) f (x) = x^{2} + 5 x + 6

\frac{( x + 4 ) ^{2}}{9} - \frac{( y - 8 ) ^{2}}{49} = 1

25 x^{2} - 9 y^{2} - 225 = 0

5 x^{2} - 4 y^{2} - 40 x + 24 y + 24 = 0

x^{2} - y^{2} - 4 x - 6 y - 14 = 0

What is the vertices 4x^2-5y^2+30y=65 ?
The vertices 4x^2-5y^2+30y=65 is (sqrt(5),3),(-sqrt(5),3)