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

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

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((x-3)^2)/(2/9)+((y-5)^2)/(25)=1

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

(h, k) = (3, 5), a = \frac{2}{3}, b = 5

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, 5), a = \frac{2}{3}, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 5), b = 5, a = \frac{2}{3}

5 x^{2} - 70 x + 9 y^{2} + 90 y + 425 = 0

(\frac{y - 3}{64})^{2} + \frac{( x + 5 ) ^{2}}{81} = 1

4 x^{2} + 9 y^{2} + 8 x + 18 y - 23 = 0

4 x^{2} + 9 y^{2} - 48 x + 72 y - 144 = 0

\frac{x ^{2}}{1 0 ^{2}} + \frac{y ^{2}}{5 ^{2}} = 1

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