What is ((x+2)^2)/4+((y-3)^2)/(16)=1 ?

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

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((x+2)^2)/4+((y-3)^2)/(16)=1

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

16 - x^{2} - 4 y^{2} = 0

25 x^{2} + 4 y^{2} - 200 x + 24 y + 376 = 0

\frac{( x + 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{25} = 1

x^{2} + y^{2} - 2 y > 0

\frac{( x - 1 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

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