What is the foci (x^2)/(81/4)+(y^2)/(25/16)=1 ?

The foci (x^2)/(81/4)+(y^2)/(25/16)=1 is ((sqrt(299))/4 ,0),(-(sqrt(299))/4 ,0)

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foci (x^2)/(81/4)+(y^2)/(25/16)=1

foci

\frac{x ^{2}}{\frac{81}{4}} + \frac{y ^{2}}{\frac{25}{16}} = 1

(\frac{299}{4}, 0), (- \frac{299}{4}, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{\frac{81}{4}} + \frac{y ^{2}}{\frac{25}{16}} = 1 :

Ellipse with center

(h, k) = (0, 0), a = \frac{9}{2}, b = \frac{5}{4}

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

Compute

c :

c = (\frac{9}{2})^{2} - (\frac{5}{4})^{2} : \frac{299}{4}

(0 + \frac{299}{4}, 0), (0 - \frac{299}{4}, 0)

Refine

(\frac{299}{4}, 0), (- \frac{299}{4}, 0)

2 x^{2} + 3 y^{2} + 4 x - 1 = 0

x^{2} + \frac{( y - 6 ) ^{2}}{16} = 1

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

\frac{( x + 5 ) ^{2}}{20} + \frac{( y - 6 ) ^{2}}{11} = 1

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 3

What is the foci (x^2)/(81/4)+(y^2)/(25/16)=1 ?
The foci (x^2)/(81/4)+(y^2)/(25/16)=1 is ((sqrt(299))/4 ,0),(-(sqrt(299))/4 ,0)