What is (x^2)/9+((y+3)^2)/(16)=1 ?

The solution to (x^2)/9+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(0,-3),a=3,b=4

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(x^2)/9+((y+3)^2)/(16)=1

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

(h, k) = (0, - 3), a = 3, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{3 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, - 3), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 3), b = 4, a = 3

16 x^{2} + 9 y^{2} - 96 x + 36 y - 36 = 0

2 x^{2} + 3 y^{2} = 10

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

9 x^{2} + 16 y^{2} + 126 x + 96 y + 441 = 0

2 x^{2} + y^{2} - 10 = 0

What is (x^2)/9+((y+3)^2)/(16)=1 ?
The solution to (x^2)/9+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(0,-3),a=3,b=4