What is the foci ((x-1)^2)/(40)+((y+4)^2)/(36)=1 ?

The foci ((x-1)^2)/(40)+((y+4)^2)/(36)=1 is (3,-4),(-1,-4)

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foci ((x-1)^2)/(40)+((y+4)^2)/(36)=1

foci

\frac{( x - 1 ) ^{2}}{40} + \frac{( y + 4 ) ^{2}}{36} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{40} + \frac{( y + 4 ) ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (1, - 4), a = 210, b = 6

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

Compute

c :

c = (210)^{2} - 6^{2} : 2

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

Refine

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

\frac{x ^{2}}{( \frac{5}{2} ) ^{2}} + \frac{y ^{2}}{( \frac{3}{2} ) ^{2}} = 1

x^{2} + 4 y^{2} - 6 x + 8 y - 3 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{121} = 1

2 x^{2} + y^{2} + 12 x - 2 y + 11 = 0

16 x^{2} + 9 y^{2} + 64 x - 80 = 0

What is the foci ((x-1)^2)/(40)+((y+4)^2)/(36)=1 ?
The foci ((x-1)^2)/(40)+((y+4)^2)/(36)=1 is (3,-4),(-1,-4)