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

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

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

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

(h, k) = (5, - 2), a = 5, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (5, - 2), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, - 2), a = 5, b = 3

\frac{( x - 7 ) ^{2}}{25} + \frac{( y + 9 ) ^{2}}{36} = 1

\frac{x ^{2}}{2. 6 ^{2}} + \frac{y ^{2}}{2.4 3 ^{2}} = 1

6 x^{2} - 30 x + 3 y^{2} + 30 y + 111 = 0

13 x^{2} + 23 y^{2} - 51 x - 19 y - 4 = 0

250 = 2 x^{2} + \frac{y ^{2}}{2}

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