What is 16x^2+36y^2+112x-108y+133=0 ?

The solution to 16x^2+36y^2+112x-108y+133=0 is Ellipse with (h,k)=(-7/2 , 3/2),a=3,b=2

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16x^2+36y^2+112x-108y+133=0

16 x^{2} + 36 y^{2} + 112 x - 108 y + 133 = 0

(h, k) = (- \frac{7}{2}, \frac{3}{2}), a = 3, b = 2

Solution steps

16 x^{2} + 36 y^{2} + 112 x - 108 y + 133 = 0

Rewrite

16 x^{2} + 36 y^{2} + 112 x - 108 y + 133 = 0

in the form of the standard ellipse equation

\frac{( x - ( - \frac{7}{2} ) ) ^{2}}{3 ^{2}} + \frac{( y - \frac{3}{2} ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{7}{2}, \frac{3}{2}), a = 3, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- \frac{7}{2}, \frac{3}{2}), a = 3, b = 2

2 x^{2} + 11 y^{2} = 11

2 x^{2} + 3 y^{2} = 11

2 x^{2} + 3 y^{2} = 17

x^{2} + 4 y^{2} \leq 1

\frac{( X - 5 ) ^{2}}{25} + \frac{( y + 7 ^{2} ) ^{2}}{16} = 1

What is 16x^2+36y^2+112x-108y+133=0 ?
The solution to 16x^2+36y^2+112x-108y+133=0 is Ellipse with (h,k)=(-7/2 , 3/2),a=3,b=2