What is ((x-3)^2)/(225)+((y+9)^2)/(121)=1 ?

The solution to ((x-3)^2)/(225)+((y+9)^2)/(121)=1 is Ellipse with (h,k)=(3,-9),a=15,b=11

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((x-3)^2)/(225)+((y+9)^2)/(121)=1

\frac{( x - 3 ) ^{2}}{225} + \frac{( y + 9 ) ^{2}}{121} = 1

(h, k) = (3, - 9), a = 15, b = 11

Solution steps

\frac{( x - 3 ) ^{2}}{225} + \frac{( y + 9 ) ^{2}}{121} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{225} + \frac{( y + 9 ) ^{2}}{121} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 9), a = 15, b = 11

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 9), a = 15, b = 11

\frac{x ^{2}}{15} + \frac{y ^{2}}{45} = 1

100 x^{2} + 4 y^{2} = 1

\frac{( x + 11 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{1 9 ^{2}} = 1

4 x^{2} + 3 y^{2} + 4 x = 1

(x - 3)^{2} + 3 (y - 2)^{2} = 2

What is ((x-3)^2)/(225)+((y+9)^2)/(121)=1 ?
The solution to ((x-3)^2)/(225)+((y+9)^2)/(121)=1 is Ellipse with (h,k)=(3,-9),a=15,b=11