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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

\frac{( x - 2 ) ^{2}}{81} + \frac{( y + 5 ) ^{2}}{64} = 1

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

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

x^{2} + (y - c)^{2} = 1

4 y^{2} + 9 x^{2} - 24 y - 72 x + 144 = 0

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