What is ((x-4)^2)/9+((y+6)^2)/(121)=1 ?

The solution to ((x-4)^2)/9+((y+6)^2)/(121)=1 is Ellipse with (h,k)=(4,-6),a=3,b=11

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((x-4)^2)/9+((y+6)^2)/(121)=1

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{121} = 1

(h, k) = (4, - 6), a = 3, b = 11

Solution steps

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{121} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 6 ) ^{2}}{121} = 1

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{3 ^{2}} + \frac{( y - ( - 6 ) ) ^{2}}{1 1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, - 6), a = 3, b = 11

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (4, - 6), b = 11, a = 3

36 x^{2} + 16 y^{2} + 180 x - 24 y + 90 = 0

2.4 x^{2} + 4 y^{2} + 40 x - 32 y - 5 = 0

25 x^{2} + 16 y^{2} + 100 x - 32 y - 284 = 0

x^{2} + 3 y^{2} - 3 x - 5 y = 0

x^{2} - 2 x + 4 y^{2} - 24 y + 33 = 0

What is ((x-4)^2)/9+((y+6)^2)/(121)=1 ?
The solution to ((x-4)^2)/9+((y+6)^2)/(121)=1 is Ellipse with (h,k)=(4,-6),a=3,b=11