What is the foci (x^2}{32}+\frac{y^2)/9 =1 ?

The foci (x^2}{32}+\frac{y^2)/9 =1 is (sqrt(23),0),(-sqrt(23),0)

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foci (x^2}{32}+\frac{y^2)/9 =1

foci

\frac{x ^{2}}{32} + \frac{y ^{2}}{9} = 1

(23, 0), (- 23, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{32} + \frac{y ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 42, b = 3

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

Compute

c :

c = (42)^{2} - 3^{2} : 23

(0 + 23, 0), (0 - 23, 0)

Refine

(23, 0), (- 23, 0)

4 x^{2} + 25 y^{2} - 24 x + 250 y + 561 = 0

14 x^{2} - 70 = - 5 y^{2}

- \frac{( y - 1 ) ^{2}}{144} - \frac{( x + 2 ) ^{2}}{9} = 1

\frac{( x - 2 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{1} = 1

x^{2} + 4 y^{2} + 10 x + 9 y - 10 = 0

What is the foci (x^2}{32}+\frac{y^2)/9 =1 ?
The foci (x^2}{32}+\frac{y^2)/9 =1 is (sqrt(23),0),(-sqrt(23),0)