What is ((x+3)^2)/(6^2)+((y+7)^2)/(2^2)=1 ?

The solution to ((x+3)^2)/(6^2)+((y+7)^2)/(2^2)=1 is Ellipse with (h,k)=(-3,-7),a=6,b=2

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((x+3)^2)/(6^2)+((y+7)^2)/(2^2)=1

\frac{( x + 3 ) ^{2}}{6 ^{2}} + \frac{( y + 7 ) ^{2}}{2 ^{2}} = 1

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

Solution steps

\frac{( x + 3 ) ^{2}}{6 ^{2}} + \frac{( y + 7 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

25 x^{(2)} - 200 x + y^{(2)} + 4 y + 379 = 0

\frac{( x - 0 ) ^{2}}{2 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

77 x^{2} + 81 y^{2} - 828 x - 324 y = 1944

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

55 x^{2} + 5 y^{2} = 4

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