What is ((x+3)^2)/3+((y-2)^2)/4 =1 ?

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

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((x+3)^2)/3+((y-2)^2)/4 =1

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

9 x^{2} - 54 x + 4 y^{2} = - 45

x^{2} + 3 y^{2} + x - 2 y - 1 = 0

(\frac{x + 7}{16})^{2} + \frac{( y + 5 ) ^{2}}{16} = 1

3 x^{2} + 2 y^{2} + 12 x - 8 y - 16 = 0

16 x^{2} + 25 y^{2} = 1.6

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