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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

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

9 x^{2} + 16 y^{2} + 180 x - 256 y - 452 = 0

9 x^{2} + 6 y^{2} = 54

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

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