What is the vertices (x^2)/(49)+(y^2)/(16)=1 ?

The vertices (x^2)/(49)+(y^2)/(16)=1 is (7,0),(-7,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x^2)/(49)+(y^2)/(16)=1

vertices

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

(7, 0), (- 7, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = 7, b = 4

(0 + 7, 0), (0 - 7, 0)

Refine

(7, 0), (- 7, 0)

f (x) = x^{2} + 2 x - 8

\frac{x ^{2}}{36} - \frac{y ^{2}}{9} = 1

y = x^{2}

f (x) = x^{2} - 4 x - 12

25 y^{2} - 36 (x - 8)^{2} = 900

What is the vertices (x^2)/(49)+(y^2)/(16)=1 ?
The vertices (x^2)/(49)+(y^2)/(16)=1 is (7,0),(-7,0)