What is the foci ((x-4)^2)/4+((y-3)^2)/9 =1 ?

The foci ((x-4)^2)/4+((y-3)^2)/9 =1 is (4,3+sqrt(5)),(4,3-sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x-4)^2)/4+((y-3)^2)/9 =1

foci

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

(4, 3 + 5), (4, 3 - 5)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(4, 3 + c), (4, 3 - c)

Compute

c :

c = 3^{2} - 2^{2} : 5

(4, 3 + 5), (4, 3 - 5)

\frac{x ^{2}}{16} + \frac{y ^{2}}{36} = 1

\frac{( x + 1 ) ^{2}}{28} + \frac{( y + 5 ) ^{2}}{23} = 1

x^{2} - 8 x + 25 y^{2} - 100 y + 91 = 0

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

2 x^{2} + y^{2} = 50

What is the foci ((x-4)^2)/4+((y-3)^2)/9 =1 ?
The foci ((x-4)^2)/4+((y-3)^2)/9 =1 is (4,3+sqrt(5)),(4,3-sqrt(5))