What is 3x^2+4y^2-28x-16y+48=0 ?

The solution to 3x^2+4y^2-28x-16y+48=0 is Ellipse with (h,k)=(14/3 ,2),a= 10/3 ,b= 5/(sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3x^2+4y^2-28x-16y+48=0

3 x^{2} + 4 y^{2} - 28 x - 16 y + 48 = 0

(h, k) = (\frac{14}{3}, 2), a = \frac{10}{3}, b = \frac{5}{3}

Solution steps

3 x^{2} + 4 y^{2} - 28 x - 16 y + 48 = 0

Rewrite

3 x^{2} + 4 y^{2} - 28 x - 16 y + 48 = 0

in the form of the standard ellipse equation

\frac{( x - \frac{14}{3} ) ^{2}}{( \frac{10}{3} ) ^{2}} + \frac{( y - 2 ) ^{2}}{( \frac{5}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (\frac{14}{3}, 2), a = \frac{10}{3}, b = \frac{5}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (\frac{14}{3}, 2), a = \frac{10}{3}, b = \frac{5}{3}

4 x^{2} + 8 x + 9 (y - 2)^{2} = 32

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

16 x^{2} + 64 x + 4 y^{2} - 8 y + 4 = 0

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{9} = 1

49 x^{2} + 4 y^{2} = 196

What is 3x^2+4y^2-28x-16y+48=0 ?
The solution to 3x^2+4y^2-28x-16y+48=0 is Ellipse with (h,k)=(14/3 ,2),a= 10/3 ,b= 5/(sqrt(3))