What is the foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1 ?

The foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1 is (4,5),(4,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1

foci

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1

(4, 5), (4, 1)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (4, 3), b = 6, a = 42

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

Compute

c :

c = 6^{2} - (42)^{2} : 2

(4, 3 + 2), (4, 3 - 2)

Refine

(4, 5), (4, 1)

\frac{( x + 3 ) ^{2}}{21} + \frac{( y - 6 ) ^{2}}{25} = 1

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

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{81} = 1

What is the foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1 ?
The foci ((x-4)^2)/(32)+((y-3)^2)/(36)=1 is (4,5),(4,1)