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

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

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

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

(h, k) = (- 2, 3), a = 1, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 3), a = 1, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 3), b = 2, a = 1

\frac{( x - 5 ) ^{2}}{47} + \frac{( y + 3 ) ^{2}}{30} = 1

4 x^{2} - 16 x + 25 y^{2} - 84 = 0

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400

8 x^{2} + 16 y^{2} = 1

\frac{( x - 8 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{1} = 1

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