What is the foci 16x^2+81y^2=81 ?

The foci 16x^2+81y^2=81 is ((sqrt(65))/4 ,0),(-(sqrt(65))/4 ,0)

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foci 16x^2+81y^2=81

foci

16 x^{2} + 81 y^{2} = 81

(\frac{65}{4}, 0), (- \frac{65}{4}, 0)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 81 y^{2} = 81 :

Ellipse with center

(h, k) = (0, 0), a = \frac{9}{4}, b = 1

(0 + c, 0), (0 - c, 0)

Compute

c :

c = (\frac{9}{4})^{2} - 1^{2} : \frac{65}{4}

(0 + \frac{65}{4}, 0), (0 - \frac{65}{4}, 0)

Refine

(\frac{65}{4}, 0), (- \frac{65}{4}, 0)

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{64} = 1

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0

What is the foci 16x^2+81y^2=81 ?
The foci 16x^2+81y^2=81 is ((sqrt(65))/4 ,0),(-(sqrt(65))/4 ,0)