What is 1x^2+3y^2+10x+10y-11=0 ?

The solution to 1x^2+3y^2+10x+10y-11=0 is Ellipse with (h,k)=(-5,-5/3),a=sqrt(133/3),b=(sqrt(133))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1x^2+3y^2+10x+10y-11=0

1 x^{2} + 3 y^{2} + 10 x + 10 y - 11 = 0

(h, k) = (- 5, - \frac{5}{3}), a = \frac{133}{3}, b = \frac{133}{3}

Solution steps

1 \cdot x^{2} + 3 y^{2} + 10 x + 10 y - 11 = 0

Rewrite

1 x^{2} + 3 y^{2} + 10 x + 10 y - 11 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 5 ) ) ^{2}}{( \frac{133}{3} ) ^{2}} + \frac{( y - ( - \frac{5}{3} ) ) ^{2}}{( \frac{133}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 5, - \frac{5}{3}), a = \frac{133}{3}, b = \frac{133}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 5, - \frac{5}{3}), a = \frac{133}{3}, b = \frac{133}{3}

16 x^{(2)} + 25 y^{(2)} = 400

16 x^{2} - 160 x + 25 y^{2} + 350 y - 1225 = 0

9 x^{2} + 16 y^{2} - 18 x - 64 y = 71

21 y^{2} + 210 y + 25 x^{2} - 100 x = 0

\frac{( x - 1 ) ^{2}}{4} + \frac{( y + 5 ) ^{2}}{16} = 1

What is 1x^2+3y^2+10x+10y-11=0 ?
The solution to 1x^2+3y^2+10x+10y-11=0 is Ellipse with (h,k)=(-5,-5/3),a=sqrt(133/3),b=(sqrt(133))/3