What is ((x-12)^2)/(289)+((y-3)^2)/(64)=1 ?

The solution to ((x-12)^2)/(289)+((y-3)^2)/(64)=1 is Ellipse with (h,k)=(12,3),a=17,b=8

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((x-12)^2)/(289)+((y-3)^2)/(64)=1

\frac{( x - 12 ) ^{2}}{289} + \frac{( y - 3 ) ^{2}}{64} = 1

(h, k) = (12, 3), a = 17, b = 8

Solution steps

\frac{( x - 12 ) ^{2}}{289} + \frac{( y - 3 ) ^{2}}{64} = 1

Rewrite

\frac{( x - 12 ) ^{2}}{289} + \frac{( y - 3 ) ^{2}}{64} = 1

in the form of the standard ellipse equation

\frac{( x - 12 ) ^{2}}{1 7 ^{2}} + \frac{( y - 3 ) ^{2}}{8 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (12, 3), a = 17, b = 8

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (12, 3), a = 17, b = 8

3 x^{2} + 6 y^{2} - 4 x + 6 y - 3 = 0

\frac{( x + 1 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1

\frac{( x - 4 ) ^{2}}{3} + \frac{y ^{2}}{9} = 1

25 x^{2} + 16 y^{2} - 100 x - 96 y - 156 = 0

\frac{x ^{2}}{25} + \frac{( y - 3 ) ^{2}}{16} = 1

What is ((x-12)^2)/(289)+((y-3)^2)/(64)=1 ?
The solution to ((x-12)^2)/(289)+((y-3)^2)/(64)=1 is Ellipse with (h,k)=(12,3),a=17,b=8