What is the foci 6.25x^2+16y^2-12.5x+64y-29.75=0 ?

The foci 6.25x^2+16y^2-12.5x+64y-29.75=0 is (4.12249…,-2),(-2.12249…,-2)

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foci 6.25x^2+16y^2-12.5x+64y-29.75=0

foci

(4.12249 \dots, - 2), (- 2.12249 \dots, - 2)

Solution steps

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

Calculate ellipse properties

6.25 x^{2} + 16 y^{2} - 12.5 x + 64 y - 29.75 = 0 :

Ellipse with center

(h, k) = (1, - 2), a = 4, b = 2.5

(1 + c, - 2), (1 - c, - 2)

Compute

c :

c = 4^{2} - 2. 5^{2} : 3.12249 \dots

(1 + 3.12249 \dots, - 2), (1 - 3.12249 \dots, - 2)

Refine

(4.12249 \dots, - 2), (- 2.12249 \dots, - 2)

What is the foci 6.25x^2+16y^2-12.5x+64y-29.75=0 ?
The foci 6.25x^2+16y^2-12.5x+64y-29.75=0 is (4.12249…,-2),(-2.12249…,-2)