What is (2-x^2-y^2)/(x^2+y^2-9)>0 ?

The solution to (2-x^2-y^2)/(x^2+y^2-9)>0 is Ellipse with (h,k)=(0,0),a=sqrt(2),b=sqrt(2)

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(2-x^2-y^2)/(x^2+y^2-9)>0

\frac{2 - x ^{2} - y ^{2}}{x ^{2} + y ^{2} - 9} > 0

(h, k) = (0, 0), a = 2, b = 2

Solution steps

\frac{2 - x ^{2} - y ^{2}}{x ^{2} + y ^{2} - 9} > 0

Rewrite

\frac{2 - x ^{2} - y ^{2}}{x ^{2} + y ^{2} - 9} > 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 2, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

f (x) = x^{2} + 3 f (x) (f (x))

x^{2} + 4 y^{2} - 6 x - 8 y + 9 = 0

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

\frac{( x + 1.23 ) ^{2}}{1.5 9 ^{2}} + \frac{y ^{2}}{1.0 1 ^{2}} = 1

\frac{x ^{2}}{12} + \frac{y ^{2}}{9} = 1

What is (2-x^2-y^2)/(x^2+y^2-9)>0 ?
The solution to (2-x^2-y^2)/(x^2+y^2-9)>0 is Ellipse with (h,k)=(0,0),a=sqrt(2),b=sqrt(2)