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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

4 x^{2} + 9 y^{2} - 24 x + 72 y + 144 = 0

4 x^{2} + 9 y^{2} + 24 x + 18 y + 9 = 0

x^{2} + 5 y^{2} = 2

x^{2} + 4 y^{2} + 2 x + 8 y + 1 = 0

25 x^{2} + 16 y^{2} + 150 x - 128 y - 1119 = 0

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