What is ((x-2}{16})^2+(\frac{y+3)/9)^2=1 ?

The solution to ((x-2}{16})^2+(\frac{y+3)/9)^2=1 is Ellipse with (h,k)=(2,-3),a=16,b=9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((x-2}{16})^2+(\frac{y+3)/9)^2=1

(\frac{x - 2}{16})^{2} + (\frac{y + 3}{9})^{2} = 1

(h, k) = (2, - 3), a = 16, b = 9

Solution steps

(\frac{x - 2}{16})^{2} + (\frac{y + 3}{9})^{2} = 1

Rewrite

(\frac{x - 2}{16})^{2} + (\frac{y + 3}{9})^{2} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{1 6 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{9 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, - 3), a = 16, b = 9

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 3), a = 16, b = 9

4 x^{2} + y^{2} - 16 x + 6 y + 9 = 0

x^{2} + 9 y^{2} = 18

9 x^{2} + 4 y^{2} - 18 x + 8 y = 23

1 x^{2} + 2 y^{2} = 9

\frac{( x - 3 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{16} = 1

What is ((x-2}{16})^2+(\frac{y+3)/9)^2=1 ?
The solution to ((x-2}{16})^2+(\frac{y+3)/9)^2=1 is Ellipse with (h,k)=(2,-3),a=16,b=9