What is ((x-4)^2)/4+((y+2)^2)/9 =1 ?

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

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((x-4)^2)/4+((y+2)^2)/9 =1

\frac{( x - 4 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

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

Solution steps

\frac{( x - 4 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

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

\frac{x ^{2}}{2} + \frac{y ^{2}}{a - 16} = 1

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

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{9} = 1

9 x^{2} + 4 y^{2} + 18 x - 27 = 0

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