What is 1x^2+4y^2+10x+8y-19=0 ?

The solution to 1x^2+4y^2+10x+8y-19=0 is Ellipse with (h,k)=(-5,-1),a=4sqrt(3),b=2sqrt(3)

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1x^2+4y^2+10x+8y-19=0

1 x^{2} + 4 y^{2} + 10 x + 8 y - 19 = 0

(h, k) = (- 5, - 1), a = 43, b = 23

Solution steps

1 \cdot x^{2} + 4 y^{2} + 10 x + 8 y - 19 = 0

Rewrite

1 x^{2} + 4 y^{2} + 10 x + 8 y - 19 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 5 ) ) ^{2}}{( 4 3 ) ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{( 2 3 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 5, - 1), a = 43, b = 23

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 5, - 1), a = 43, b = 23

36 x^{2} + 9 y^{2} + 288 x - 90 y + 477 = 0

(\frac{x ^{2}}{72}) + (\frac{y ^{2}}{21}) = 1

(3 - y^{2} - 3 x^{2}) = 0

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 7 ) ^{2}}{36} = 1

\frac{( ( x - 2 ) ^{2} )}{( 9 )} + \frac{( ( y + 5 ) ^{2} )}{( 25 )} = 1

What is 1x^2+4y^2+10x+8y-19=0 ?
The solution to 1x^2+4y^2+10x+8y-19=0 is Ellipse with (h,k)=(-5,-1),a=4sqrt(3),b=2sqrt(3)