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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

\frac{x ^{2} + y ^{2}}{x} = 16

4 x^{(2)} + 16 y^{(2)} - 16 x - 48 = 0

\frac{( x ^{2} )}{( 16 )} + \frac{( y ^{2} )}{( 12 )} = 1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 7 ) ^{2}}{64} = 1

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