What is the foci ((x+1)^2)/9+((y-3)^2)/(10)=1 ?

The foci ((x+1)^2)/9+((y-3)^2)/(10)=1 is (-1,4),(-1,2)

Popular >

foci ((x+1)^2)/9+((y-3)^2)/(10)=1

foci

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{10} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{10} = 1 :

Ellipse with center

(h, k) = (- 1, 3), b = 10, a = 3

(- 1, 3 + c), (- 1, 3 - c)

Compute

c :

c = 10^{2} - 3^{2} : 1

(- 1, 3 + 1), (- 1, 3 - 1)

Refine

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

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

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{4} = 1

4 x^{2} + 9 y^{2} + 12 x - 108 y + 297 = 0

9 x^{(2)} + 16 y^{(2)} - 54 x + 64 y + 1 = 0

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

What is the foci ((x+1)^2)/9+((y-3)^2)/(10)=1 ?
The foci ((x+1)^2)/9+((y-3)^2)/(10)=1 is (-1,4),(-1,2)