What is the center 3x^2+y^2-24x+39=0 ?

The center 3x^2+y^2-24x+39=0 is (4,0)

Popular >

center 3x^2+y^2-24x+39=0

center

3 x^{2} + y^{2} - 24 x + 39 = 0

(4, 0)

Solution steps

3 x^{2} + y^{2} - 24 x + 39 = 0

Rewrite

3 x^{2} + y^{2} - 24 x + 39 = 0

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{( 3 ) ^{2}} + \frac{( y - 0 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, 0), a = 3, b = 3

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (4, 0), b = 3, a = 3

And the center is:

(4, 0)

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

64 x^{2} + 100 y^{2} + 768 x - 600 y - 3196 = 0

2 x^{2} + 3 y^{2} - 4 x + 6 y - 1 = 0

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

4 x^{2} + 25 y^{2} = 100