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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

25 x^{2} + 4 y^{2} - 50 x + 16 y - 59 = 0

5 x^{2} + 7 y^{2} = 35

(\frac{x ^{2}}{a}) + (\frac{y ^{2}}{b}) = 1

\frac{( x - 10 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{36} = 1

3 x^{2} + 2 y^{2} - 6 x + 4 y - 2 = 0

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