What is the foci ((x+4)^2)/(25)+((y-1)^2)/(16)=1 ?

The foci ((x+4)^2)/(25)+((y-1)^2)/(16)=1 is (-1,1),(-7,1)

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foci ((x+4)^2)/(25)+((y-1)^2)/(16)=1

foci

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1

(- 1, 1), (- 7, 1)

Solution steps

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

Calculate ellipse properties

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (- 4, 1), a = 5, b = 4

(- 4 + c, 1), (- 4 - c, 1)

Compute

c :

c = 5^{2} - 4^{2} : 3

(- 4 + 3, 1), (- 4 - 3, 1)

Refine

(- 1, 1), (- 7, 1)

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

\frac{x ^{2}}{81} + \frac{y ^{2}}{100} = 1

x^{2} + 4 y^{2} - 6 x + 16 y + 21 = 0

\frac{x ^{2}}{64} + \frac{y ^{2}}{15} = 1

4 x^{2} + 24 x + 25 y^{2} + 200 y + 336 = 0

What is the foci ((x+4)^2)/(25)+((y-1)^2)/(16)=1 ?
The foci ((x+4)^2)/(25)+((y-1)^2)/(16)=1 is (-1,1),(-7,1)