What is the foci 16x^2+25y^2-64x+150y=-279 ?

The foci 16x^2+25y^2-64x+150y=-279 is ((4sqrt(10)+3)/(2sqrt(10)),-3),((4sqrt(10)-3)/(2sqrt(10)),-3)

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foci 16x^2+25y^2-64x+150y=-279

foci

16 x^{2} + 25 y^{2} - 64 x + 150 y = - 279

(\frac{4 10 + 3}{2 10}, - 3), (\frac{4 10 - 3}{2 10}, - 3)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} - 64 x + 150 y = - 279 :

Ellipse with center

(h, k) = (2, - 3), a = \frac{5}{2 2}, b = \frac{2}{5}

(2 + c, - 3), (2 - c, - 3)

Compute

c :

c = (\frac{5}{2 2})^{2} - \frac{2}{5}^{2} : \frac{3}{2 10}

(2 + \frac{3}{2 10}, - 3), (2 - \frac{3}{2 10}, - 3)

Refine

(\frac{4 10 + 3}{2 10}, - 3), (\frac{4 10 - 3}{2 10}, - 3)

\frac{( x - 12 ) ^{2}}{100} + \frac{( y + 7 ) ^{2}}{36} = 1

\frac{x ^{2}}{0.38 4 ^{2}} + \frac{y ^{2}}{0.375 8 ^{2}} = 1

4 x^{2} + y^{2} + 6 y + 5 = 0

9 (x + 2)^{2} + (y + 1)^{2} = 1

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

What is the foci 16x^2+25y^2-64x+150y=-279 ?
The foci 16x^2+25y^2-64x+150y=-279 is ((4sqrt(10)+3)/(2sqrt(10)),-3),((4sqrt(10)-3)/(2sqrt(10)),-3)