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

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

25 xx + 9 yy - 18 y - 216 = 0

2 x^{2} + 5 y^{2} = 14

5 x^{2} + y^{2} + x + 8 y + 5 = 0

\frac{( x - 5 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{36} = 1

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