What is the vertices-4x^2+24x+16y^2-128y+156=0 ?

The vertices-4x^2+24x+16y^2-128y+156=0 is (3,6),(3,2)

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

vertices

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

(3, 6), (3, 2)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (3, 4), a = 2, b = 4

(3, 4 + 2), (3, 4 - 2)

Refine

(3, 6), (3, 2)

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

4 x^{2} - 8 y^{2} = 3

2 x^{2} - 3 y^{2} = 6

What is the vertices-4x^2+24x+16y^2-128y+156=0 ?
The vertices-4x^2+24x+16y^2-128y+156=0 is (3,6),(3,2)