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

The foci (x^2)/(49)+(y^2)/(16)=1 is (sqrt(33),0),(-sqrt(33),0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

foci

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

(33, 0), (- 33, 0)

Solution steps

(h + c, k), (h - c, 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 + c, 0), (0 - c, 0)

Compute

c :

c = 7^{2} - 4^{2} : 33

(0 + 33, 0), (0 - 33, 0)

Refine

(33, 0), (- 33, 0)

9 x^{2} + 16 y^{2} - 144 = 0

x^{2} + y^{2} = 49

f (x) = x^{2} - 1

36 x^{2} + 36 y^{2} - 12 x - 27 y - 8 = 0

y^{2} = 2 x

What is the foci (x^2)/(49)+(y^2)/(16)=1 ?
The foci (x^2)/(49)+(y^2)/(16)=1 is (sqrt(33),0),(-sqrt(33),0)