What is ((2x+1)^2)/4+((3y-2)^2)/(16)=1 ?

The solution to ((2x+1)^2)/4+((3y-2)^2)/(16)=1 is Ellipse with (h,k)=(-1/2 , 2/3),a=1,b= 4/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((2x+1)^2)/4+((3y-2)^2)/(16)=1

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

(h, k) = (- \frac{1}{2}, \frac{2}{3}), a = 1, b = \frac{4}{3}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- \frac{1}{2}, \frac{2}{3}), a = 1, b = \frac{4}{3}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- \frac{1}{2}, \frac{2}{3}), b = \frac{4}{3}, a = 1

\frac{( x + 1 ) ^{2}}{100} + \frac{( y + 2 ) ^{2}}{121} = 1

25 x^{2} + 16 y^{2} < 400

x^{2} + y^{2} = 6 y - 8 y

25 x^{2} + 49 y^{2} = 1225

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

What is ((2x+1)^2)/4+((3y-2)^2)/(16)=1 ?
The solution to ((2x+1)^2)/4+((3y-2)^2)/(16)=1 is Ellipse with (h,k)=(-1/2 , 2/3),a=1,b= 4/3