What is ((x-5)^2}{25}+\frac{(y+3)^2)/9 =1 ?

The solution to ((x-5)^2}{25}+\frac{(y+3)^2)/9 =1 is Ellipse with (h,k)=(5,-3),a=5,b=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((x-5)^2}{25}+\frac{(y+3)^2)/9 =1

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{9} = 1

(h, k) = (5, - 3), a = 5, b = 3

Solution steps

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (5, - 3), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, - 3), a = 5, b = 3

16 y^{2} - 25 x^{2} = 400

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

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

\frac{x ^{2}}{64} + \frac{y ^{2}}{36} = 1

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

What is ((x-5)^2}{25}+\frac{(y+3)^2)/9 =1 ?
The solution to ((x-5)^2}{25}+\frac{(y+3)^2)/9 =1 is Ellipse with (h,k)=(5,-3),a=5,b=3