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

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

Popular >

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

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

(h, k) = (5, 3), a = 2, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (5, 3), a = 2, b = 3

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, 3), b = 3, a = 2

25 x^{2} + 9 y^{2} + 100 x - 125 = 0

\frac{1}{x + i y} + \frac{1}{x - i y} < - 1

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

6 x^{2} + 9 y^{2} - 24 x - 54 y + 51 = 0

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

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