What is the foci 4x^2+9y^2-8x-32=0 ?

The foci 4x^2+9y^2-8x-32=0 is (1+sqrt(5),0),(1-sqrt(5),0)

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foci 4x^2+9y^2-8x-32=0

foci

4 x^{2} + 9 y^{2} - 8 x - 32 = 0

(1 + 5, 0), (1 - 5, 0)

Solution steps

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

Calculate ellipse properties

4 x^{2} + 9 y^{2} - 8 x - 32 = 0 :

Ellipse with center

(h, k) = (1, 0), a = 3, b = 2

(1 + c, 0), (1 - c, 0)

Compute

c :

c = 3^{2} - 2^{2} : 5

(1 + 5, 0), (1 - 5, 0)

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1

\frac{( x + 3 ) ^{2}}{21} + \frac{( y - 6 ) ^{2}}{25} = 1

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

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1

\frac{x ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

What is the foci 4x^2+9y^2-8x-32=0 ?
The foci 4x^2+9y^2-8x-32=0 is (1+sqrt(5),0),(1-sqrt(5),0)