What is ((x-2)^2)/5+((y+1)^2)/4 =1 ?

The solution to ((x-2)^2)/5+((y+1)^2)/4 =1 is Ellipse with (h,k)=(2,-1),a=sqrt(5),b=2

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((x-2)^2)/5+((y+1)^2)/4 =1

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

(h, k) = (2, - 1), a = 5, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 1), a = 5, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 1), a = 5, b = 2

6 x^{2} + y^{2} + 12 x = 0

2 - \frac{2 - 2 x}{2 x - x ^{2} - y ^{2}} = 0

(\frac{x ^{2}}{64}) + (\frac{y ^{2}}{36}) = 1

X^{2} + 2 \cdot Y^{2} - 2 \cdot X - 4 \cdot Y - 8 = 0

\frac{x ^{2}}{21} + \frac{y ^{2}}{25} = 1

What is ((x-2)^2)/5+((y+1)^2)/4 =1 ?
The solution to ((x-2)^2)/5+((y+1)^2)/4 =1 is Ellipse with (h,k)=(2,-1),a=sqrt(5),b=2