What is 4x^2+25y^2+4x-120y+45=0 ?

The solution to 4x^2+25y^2+4x-120y+45=0 is Ellipse with (h,k)=(-1/2 , 12/5),a=5,b=2

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4x^2+25y^2+4x-120y+45=0

4 x^{2} + 25 y^{2} + 4 x - 120 y + 45 = 0

(h, k) = (- \frac{1}{2}, \frac{12}{5}), a = 5, b = 2

Solution steps

4 x^{2} + 25 y^{2} + 4 x - 120 y + 45 = 0

Rewrite

4 x^{2} + 25 y^{2} + 4 x - 120 y + 45 = 0

in the form of the standard ellipse equation

\frac{( x - ( - \frac{1}{2} ) ) ^{2}}{5 ^{2}} + \frac{( y - \frac{12}{5} ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{1}{2}, \frac{12}{5}), a = 5, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- \frac{1}{2}, \frac{12}{5}), a = 5, b = 2

4 x^{2} + y^{2} = 72

64 x^{2} + 25 y^{2} - 16 x - 16 y - 648 = 0

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

3 x^{2} + 4 y^{2} = 12

25 x^{2} + 16 y^{2} - 150 x + 128 y + 81 = 0

What is 4x^2+25y^2+4x-120y+45=0 ?
The solution to 4x^2+25y^2+4x-120y+45=0 is Ellipse with (h,k)=(-1/2 , 12/5),a=5,b=2