What is the vertices ((x+3)^2)/4-((y-1)^2)/(16)=1 ?

The vertices ((x+3)^2)/4-((y-1)^2)/(16)=1 is (-1,1),(-5,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices ((x+3)^2)/4-((y-1)^2)/(16)=1

vertices

\frac{( x + 3 ) ^{2}}{4} - \frac{( y - 1 ) ^{2}}{16} = 1

(- 1, 1), (- 5, 1)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 3 ) ^{2}}{4} - \frac{( y - 1 ) ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (- 3, 1), a = 2, b = 4

(- 3 + 2, 1), (- 3 - 2, 1)

Refine

(- 1, 1), (- 5, 1)

(\frac{y - 4}{16})^{2} - (\frac{x - 6}{16})^{2} = 1

\frac{( x - 1 ) ^{2}}{25} - \frac{( y + 6 ) ^{2}}{49} = 1

4 x^{2} - 24 x - 36 y^{2} - 360 y + 864 = 0

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

\frac{( y - 6 ) ^{2}}{36} - \frac{( x + 1 ) ^{2}}{16} = 1

What is the vertices ((x+3)^2)/4-((y-1)^2)/(16)=1 ?
The vertices ((x+3)^2)/4-((y-1)^2)/(16)=1 is (-1,1),(-5,1)