What is x^2+4y^2+10x+8y-13=0 ?

The solution to x^2+4y^2+10x+8y-13=0 is Ellipse with (h,k)=(-5,-1),a=sqrt(42),b=sqrt(21/2)

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x^2+4y^2+10x+8y-13=0

x^{2} + 4 y^{2} + 10 x + 8 y - 13 = 0

(h, k) = (- 5, - 1), a = 42, b = \frac{21}{2}

Solution steps

x^{2} + 4 y^{2} + 10 x + 8 y - 13 = 0

Rewrite

x^{2} + 4 y^{2} + 10 x + 8 y - 13 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 5 ) ) ^{2}}{( 42 ) ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{( \frac{21}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 5, - 1), a = 42, b = \frac{21}{2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 5, - 1), a = 42, b = \frac{21}{2}

25 x^{2} + 4 y^{2} - 100 x + 8 y + 91 = 0

4 x^{2} + 9 y^{2} - 54 y + 45 = 0

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

3 x^{2} + 4 y^{2} + 18 x + 8 y + 19 = 0

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

What is x^2+4y^2+10x+8y-13=0 ?
The solution to x^2+4y^2+10x+8y-13=0 is Ellipse with (h,k)=(-5,-1),a=sqrt(42),b=sqrt(21/2)