What is 2x^2+4y^2+10x+9y-5=0 ?

The solution to 2x^2+4y^2+10x+9y-5=0 is Ellipse with (h,k)=(-5/2 ,-9/8),a=(19)/(4sqrt(2)),b= 19/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^2+4y^2+10x+9y-5=0

2 x^{2} + 4 y^{2} + 10 x + 9 y - 5 = 0

(h, k) = (- \frac{5}{2}, - \frac{9}{8}), a = \frac{19}{4 2}, b = \frac{19}{8}

Solution steps

2 x^{2} + 4 y^{2} + 10 x + 9 y - 5 = 0

Rewrite

2 x^{2} + 4 y^{2} + 10 x + 9 y - 5 = 0

in the form of the standard ellipse equation

\frac{( x - ( - \frac{5}{2} ) ) ^{2}}{( \frac{19}{4 2} ) ^{2}} + \frac{( y - ( - \frac{9}{8} ) ) ^{2}}{( \frac{19}{8} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{5}{2}, - \frac{9}{8}), a = \frac{19}{4 2}, b = \frac{19}{8}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- \frac{5}{2}, - \frac{9}{8}), a = \frac{19}{4 2}, b = \frac{19}{8}

\frac{( y - 6 ) ^{2}}{36} + \frac{( x + 3 ) ^{2}}{16} = 1

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

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

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

9 x^{2} + 4 y^{2} - 126 x - 40 y + 505 = 0

What is 2x^2+4y^2+10x+9y-5=0 ?
The solution to 2x^2+4y^2+10x+9y-5=0 is Ellipse with (h,k)=(-5/2 ,-9/8),a=(19)/(4sqrt(2)),b= 19/8