What is the vertices-4x^2+y^2+16x+12y+16=0 ?

The vertices-4x^2+y^2+16x+12y+16=0 is (2,-4),(2,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices-4x^2+y^2+16x+12y+16=0

vertices

- 4 x^{2} + y^{2} + 16 x + 12 y + 16 = 0

(2, - 4), (2, - 8)

Solution steps

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

Calculate Hyperbola properties

- 4 x^{2} + y^{2} + 16 x + 12 y + 16 = 0 :

Up-down Hyperbola with

(h, k) = (2, - 6), a = 2, b = 1

(2, - 6 + 2), (2, - 6 - 2)

Refine

(2, - 4), (2, - 8)

\frac{y ^{2}}{81} - \frac{x ^{2}}{64} = 1

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

4 (y - 3)^{2} - 9 (x - 2)^{2} = 36

\frac{x ^{2}}{36} - \frac{( y - 10 ) ^{2}}{64} = 1

4 x^{2} - y^{2} + 32 x + 10 y + 38 = 0

What is the vertices-4x^2+y^2+16x+12y+16=0 ?
The vertices-4x^2+y^2+16x+12y+16=0 is (2,-4),(2,-8)