What is the center 4x^2+y^2-5=0 ?

The center 4x^2+y^2-5=0 is (0,0)

Popular >

center 4x^2+y^2-5=0

center

4 x^{2} + y^{2} - 5 = 0

(0, 0)

Solution steps

4 x^{2} + y^{2} - 5 = 0

Rewrite

4 x^{2} + y^{2} - 5 = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{5}{2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( 5 ) ^{2}} = 1

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 5, a = \frac{5}{2}

And the center is:

(0, 0)

\frac{x ^{2}}{81} + \frac{y ^{2}}{64} = 1

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{9} = 1

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{36} = 1

\frac{( x - 3 ) ^{2}}{86} + \frac{( y + 5 ) ^{2}}{49} = 1

x^{2} + y^{2} - 6 y = 0