What is x^2+5y^2+10y+4x-27=0 ?

The solution to x^2+5y^2+10y+4x-27=0 is Ellipse with (h,k)=(-2,-1),a=6,b= 6/(sqrt(5))

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x^2+5y^2+10y+4x-27=0

x^{2} + 5 y^{2} + 10 y + 4 x - 27 = 0

(h, k) = (- 2, - 1), a = 6, b = \frac{6}{5}

Solution steps

x^{2} + 5 y^{2} + 10 y + 4 x - 27 = 0

Rewrite

x^{2} + 5 y^{2} + 10 y + 4 x - 27 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 2 ) ) ^{2}}{6 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{( \frac{6}{5} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, - 1), a = 6, b = \frac{6}{5}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, - 1), a = 6, b = \frac{6}{5}

x^{2} + 4 y^{2} \geq 100

4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

9 x^{2} + 6 y^{2} + 36 y = 0

\frac{x ^{2}}{a} + \frac{y ^{2}}{b} = 1

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

What is x^2+5y^2+10y+4x-27=0 ?
The solution to x^2+5y^2+10y+4x-27=0 is Ellipse with (h,k)=(-2,-1),a=6,b= 6/(sqrt(5))