What is ((x-5)^2}{25}+\frac{(y+4)^2)/4 =1 ?

The solution to ((x-5)^2}{25}+\frac{(y+4)^2)/4 =1 is Ellipse with (h,k)=(5,-4),a=5,b=2

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+4)^2)/4 =1

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 4 ) ^{2}}{4} = 1

(h, k) = (5, - 4), a = 5, b = 2

Solution steps

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 4 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 4 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (5, - 4), a = 5, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, - 4), a = 5, b = 2

5 x - 2 x^{2} - 5 y - 5 y^{2} + 8 = 0

\frac{( x - 1 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{36} = 1

9 x^{2} + 16 y^{2} = 1

16 x^{2} + 9 y^{2} - 144 = 0

2 x^{2} + 3 y^{2} = 16

What is ((x-5)^2}{25}+\frac{(y+4)^2)/4 =1 ?
The solution to ((x-5)^2}{25}+\frac{(y+4)^2)/4 =1 is Ellipse with (h,k)=(5,-4),a=5,b=2