What is ((X+3)^2)/(49)+((Y+2)^2)/(36)=1 ?

The solution to ((X+3)^2)/(49)+((Y+2)^2)/(36)=1 is Ellipse with (h,k)=(-3,-2),a=7,b=6

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((X+3)^2)/(49)+((Y+2)^2)/(36)=1

\frac{( X + 3 ) ^{2}}{49} + \frac{( Y + 2 ) ^{2}}{36} = 1

(h, k) = (- 3, - 2), a = 7, b = 6

Solution steps

\frac{( X + 3 ) ^{2}}{49} + \frac{( Y + 2 ) ^{2}}{36} = 1

Rewrite

\frac{( X + 3 ) ^{2}}{49} + \frac{( Y + 2 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( X - ( - 3 ) ) ^{2}}{7 ^{2}} + \frac{( Y - ( - 2 ) ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, - 2), a = 7, b = 6

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 3, - 2), a = 7, b = 6

16 x^{2} + 25 y^{2} + 224 + 250 y - 191 = 0

\frac{( x + 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{9} = 1

9 y^{2} + 36 y + 4 x^{2} = 0

16 x^{(2)} + 25 y^{(2)} - 128 x + 100 y = 0

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 8 ) ^{2}}{36} = 1

What is ((X+3)^2)/(49)+((Y+2)^2)/(36)=1 ?
The solution to ((X+3)^2)/(49)+((Y+2)^2)/(36)=1 is Ellipse with (h,k)=(-3,-2),a=7,b=6