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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

\frac{100}{1 + x ^{2} + 2 y ^{2}} = 1

81 x^{2} + 49 y^{2} + 486 x - 490 y - 2015 = 0

0 = 1 + 2 x^{2} + 2 y^{2}

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

16 x^{2} + 9 y^{2} - 32 x + 54 y - 47 = 0

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