What is 10-(x^2)/2-y^2=x^2+4y^2 ?

The solution to 10-(x^2)/2-y^2=x^2+4y^2 is Ellipse with (h,k)=(0,0),a=(2sqrt(5))/(sqrt(3)),b=sqrt(2)

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10-(x^2)/2-y^2=x^2+4y^2

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

(h, k) = (0, 0), a = \frac{2 5}{3}, b = 2

Solution steps

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

Rewrite

10 - \frac{x ^{2}}{2} - y^{2} - x^{2} - 4 y^{2} = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{2 5}{3} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( 2 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{2 5}{3}, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = \frac{2 5}{3}, b = 2

x^{2} + 1 - 2 x + 4 y^{2} + 16 y > 0

(x + 5)^{2} + 9 (y + 3)^{2} = 36

\frac{( x + 5 ) ^{2}}{4} + \frac{( y + 5 ) ^{2}}{25} = 1

x^{2} + 4 y^{2} = 15

2 x^{2} + y^{2} - 8 y - 2 x + 1 = 0

What is 10-(x^2)/2-y^2=x^2+4y^2 ?
The solution to 10-(x^2)/2-y^2=x^2+4y^2 is Ellipse with (h,k)=(0,0),a=(2sqrt(5))/(sqrt(3)),b=sqrt(2)