What is ((x-5)^2}{25}+\frac{(y-4)^2)/9 =1 ?

The solution to ((x-5)^2}{25}+\frac{(y-4)^2)/9 =1 is Ellipse with (h,k)=(5,4),a=5,b=3

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((x-5)^2}{25}+\frac{(y-4)^2)/9 =1

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{9} = 1

(h, k) = (5, 4), a = 5, b = 3

Solution steps

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{5 ^{2}} + \frac{( y - 4 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, 4), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, 4), a = 5, b = 3

9 x^{2} + 4 y^{2} + 18 x - 27 = 0

3 x^{2} + 4 y^{2} - 18 x - 8 y - 17 = 0

\frac{x ^{2}}{20} + \frac{y ^{2}}{16} = 1

\frac{( x + 7 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{25} = 1

49 x^{2} + y^{2} - 14 y = 0

What is ((x-5)^2}{25}+\frac{(y-4)^2)/9 =1 ?
The solution to ((x-5)^2}{25}+\frac{(y-4)^2)/9 =1 is Ellipse with (h,k)=(5,4),a=5,b=3