What is the vertices 2x^2-2y^2-20x+16y+16=0 ?

The vertices 2x^2-2y^2-20x+16y+16=0 is (6,4),(4,4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 2x^2-2y^2-20x+16y+16=0

vertices

2 x^{2} - 2 y^{2} - 20 x + 16 y + 16 = 0

(6, 4), (4, 4)

Solution steps

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

Calculate Hyperbola properties

2 x^{2} - 2 y^{2} - 20 x + 16 y + 16 = 0 :

Right-left Hyperbola with

(h, k) = (5, 4), a = 1, b = 1

(5 + 1, 4), (5 - 1, 4)

Refine

(6, 4), (4, 4)

\frac{x ^{2}}{144} - \frac{y ^{2}}{25} = 1

9 y^{2} - x^{2} + 2 x + 54 y + 62 = 0

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 5 ) ^{2}}{9} = 1

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0

x^{2} - y^{2} + 2 y = 0

What is the vertices 2x^2-2y^2-20x+16y+16=0 ?
The vertices 2x^2-2y^2-20x+16y+16=0 is (6,4),(4,4)