What is ((x+1)^2)/8+((y-3/2)^2)/4 =1 ?

The solution to ((x+1)^2)/8+((y-3/2)^2)/4 =1 is Ellipse with (h,k)=(-1, 3/2),a=2sqrt(2),b=2

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((x+1)^2)/8+((y-3/2)^2)/4 =1

\frac{( x + 1 ) ^{2}}{8} + \frac{( y - \frac{3}{2} ) ^{2}}{4} = 1

(h, k) = (- 1, \frac{3}{2}), a = 22, b = 2

Solution steps

\frac{( x + 1 ) ^{2}}{8} + \frac{( y - \frac{3}{2} ) ^{2}}{4} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{8} + \frac{( y - \frac{3}{2} ) ^{2}}{4} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, \frac{3}{2}), a = 22, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, \frac{3}{2}), a = 22, b = 2

\frac{x ^{2}}{1.5869} + \frac{y ^{2}}{0.9988} = 1

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

4 x^{2} + y^{2} \leq 4

x^{2} + (y - 1)^{2} + (4 + y)^{2} = 64

\frac{( y + 1 ) ^{2}}{16} + \frac{x ^{2}}{4} - 1 = 0

What is ((x+1)^2)/8+((y-3/2)^2)/4 =1 ?
The solution to ((x+1)^2)/8+((y-3/2)^2)/4 =1 is Ellipse with (h,k)=(-1, 3/2),a=2sqrt(2),b=2