What is the foci ((x+3)^2)/(49)+((y-1)^2)/(48)=1 ?

The foci ((x+3)^2)/(49)+((y-1)^2)/(48)=1 is (-2,1),(-4,1)

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foci ((x+3)^2)/(49)+((y-1)^2)/(48)=1

foci

\frac{( x + 3 ) ^{2}}{49} + \frac{( y - 1 ) ^{2}}{48} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{49} + \frac{( y - 1 ) ^{2}}{48} = 1 :

Ellipse with center

(h, k) = (- 3, 1), a = 7, b = 43

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

Compute

c :

c = 7^{2} - (43)^{2} : 1

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

Refine

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

36 x^{2} + 9 y^{2} = 324

\frac{( x + 2 ) ^{2}}{100} + \frac{( y - 5 ) ^{2}}{36} = 1

25 x^{2} + 4 y^{2} - 100 x = 0

\frac{x ^{2}}{7} + \frac{y ^{2}}{43} = 1

3 x^{2} + 10 y^{2} = 30

What is the foci ((x+3)^2)/(49)+((y-1)^2)/(48)=1 ?
The foci ((x+3)^2)/(49)+((y-1)^2)/(48)=1 is (-2,1),(-4,1)