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

The solution to ((x+1)^2)/2+(y^2)/8 =1 is Ellipse with (h,k)=(-1,0),a=sqrt(2),b=2sqrt(2)

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((x+1)^2)/2+(y^2)/8 =1

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

(h, k) = (- 1, 0), a = 2, b = 22

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = 2, b = 22

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), b = 22, a = 2

y^{2} = - 0.25 x^{2} + 1

x, \frac{x ^{2}}{25} + \frac{Y ( x ) ^{2}}{4} = 1

\frac{( x - 4 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

\frac{( x + 6.5 ) ^{2}}{56.25} + \frac{( y - 3 ) ^{2}}{36} = 1

x^{2} - 4 x + 4 + y^{2} = \frac{x ^{2}}{4} - \frac{16 x}{4} + \frac{64}{4}

What is ((x+1)^2)/2+(y^2)/8 =1 ?
The solution to ((x+1)^2)/2+(y^2)/8 =1 is Ellipse with (h,k)=(-1,0),a=sqrt(2),b=2sqrt(2)