What is ((x-3)^2)/9+((y-6)^2)/(16)=1 ?

The solution to ((x-3)^2)/9+((y-6)^2)/(16)=1 is Ellipse with (h,k)=(3,6),a=3,b=4

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((x-3)^2)/9+((y-6)^2)/(16)=1

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 6 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 6 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 6 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

5 x^{2} + 20 y^{2} - 25 x - 10 y = 5

2 x^{2} + 5 y^{2} + 8 x - 10 y - 27 = 0

\frac{( x - 4 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{20} = 1

\frac{x ^{2}}{13} + \frac{y ^{2}}{9} = 1

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{18} = 1

What is ((x-3)^2)/9+((y-6)^2)/(16)=1 ?
The solution to ((x-3)^2)/9+((y-6)^2)/(16)=1 is Ellipse with (h,k)=(3,6),a=3,b=4