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

The vertices ((x-1)^2)/4+((y+2)^2)/(16)=1 is (1,2),(1,-6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

vertices

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

(1, 2), (1, - 6)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

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

Ellipse with center

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

(1, - 2 + 4), (1, - 2 - 4)

Refine

(1, 2), (1, - 6)

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1

4 x^{2} + 9 y^{2} - 32 x - 36 y + 64 = 0

x^{2} + 16 y^{2} = 1

2 x^{2} + 7 y^{2} = 14

16 x^{2} + 25 y^{2} - 1600 = 0

What is the vertices ((x-1)^2)/4+((y+2)^2)/(16)=1 ?
The vertices ((x-1)^2)/4+((y+2)^2)/(16)=1 is (1,2),(1,-6)