What is ((x-3)^2)/8+((y+1)^2)/5 =1 ?

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

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((x-3)^2)/8+((y+1)^2)/5 =1

\frac{( x - 3 ) ^{2}}{8} + \frac{( y + 1 ) ^{2}}{5} = 1

(h, k) = (3, - 1), a = 22, b = 5

Solution steps

\frac{( x - 3 ) ^{2}}{8} + \frac{( y + 1 ) ^{2}}{5} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{8} + \frac{( y + 1 ) ^{2}}{5} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 22, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), a = 22, b = 5

\frac{( x ^{(2)} )}{( 36 )} + \frac{( y ^{(2)} )}{( 16 )} = 1

16 x^{2} + 36 y^{2} - 256 x + 576 y + 2752 = 0

16 x^{2} + 4 y^{2} + 32 x + 32 y + 16 = 0

x^{2} + 5 y^{2} = 150

36 x^{2} + 16 y^{2} = 49

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