What is ((x-2)^2}{12}+\frac{(y+5)^2)/8 =1 ?

The solution to ((x-2)^2}{12}+\frac{(y+5)^2)/8 =1 is Ellipse with (h,k)=(2,-5),a=2sqrt(3),b=2sqrt(2)

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((x-2)^2}{12}+\frac{(y+5)^2)/8 =1

\frac{( x - 2 ) ^{2}}{12} + \frac{( y + 5 ) ^{2}}{8} = 1

(h, k) = (2, - 5), a = 23, b = 22

Solution steps

\frac{( x - 2 ) ^{2}}{12} + \frac{( y + 5 ) ^{2}}{8} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{12} + \frac{( y + 5 ) ^{2}}{8} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{( 2 3 ) ^{2}} + \frac{( y - ( - 5 ) ) ^{2}}{( 2 2 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, - 5), a = 23, b = 22

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 5), a = 23, b = 22

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

4 (x - 3)^{2} + 2 y^{2} = 8

9 x^{2} + 36 y^{2} + 18 x - 360 y + 585 = 0

\frac{x ^{2}}{49} + \frac{y ^{2}}{24} = 1

x^{(2)} + 4 y^{(2)} - 6 x + 16 y + 21 = 0

What is ((x-2)^2}{12}+\frac{(y+5)^2)/8 =1 ?
The solution to ((x-2)^2}{12}+\frac{(y+5)^2)/8 =1 is Ellipse with (h,k)=(2,-5),a=2sqrt(3),b=2sqrt(2)