What is ((x+3)^2)/(12)+((y-2)^2)/(16)=1 ?

The solution to ((x+3)^2)/(12)+((y-2)^2)/(16)=1 is Ellipse with (h,k)=(-3,2),a=2sqrt(3),b=4

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((x+3)^2)/(12)+((y-2)^2)/(16)=1

\frac{( x + 3 ) ^{2}}{12} + \frac{( y - 2 ) ^{2}}{16} = 1

(h, k) = (- 3, 2), a = 23, b = 4

Solution steps

\frac{( x + 3 ) ^{2}}{12} + \frac{( y - 2 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{12} + \frac{( y - 2 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 3, 2), a = 23, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, 2), b = 4, a = 23

2 x^{2} + y^{2} - 2 x + 3 y = 5

\frac{x ^{2}}{81} + \frac{y ^{2}}{100} = 1

25 (x + 1)^{2} + 16 (y - 3)^{2} = 400

9 y^{2} + 90 y + 4 x^{2} - 16 x + 205 = 0

49 x^{2} + 81 y^{2} = 3969

What is ((x+3)^2)/(12)+((y-2)^2)/(16)=1 ?
The solution to ((x+3)^2)/(12)+((y-2)^2)/(16)=1 is Ellipse with (h,k)=(-3,2),a=2sqrt(3),b=4