What is 9= 1/(x^2+y^2-7) ?

The solution to 9= 1/(x^2+y^2-7) is Ellipse with (h,k)=(0,0),a= 8/3 ,b= 8/3

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9= 1/(x^2+y^2-7)

9 = \frac{1}{x ^{2} + y ^{2} - 7}

(h, k) = (0, 0), a = \frac{8}{3}, b = \frac{8}{3}

Solution steps

9 = \frac{1}{x ^{2} + y ^{2} - 7}

Rewrite

9 - \frac{1}{x ^{2} + y ^{2} - 7} = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{8}{3} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{8}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{8}{3}, b = \frac{8}{3}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = \frac{8}{3}, a = \frac{8}{3}

x^{2} + 8 y^{2} - 12 x - 16 y - 36 = 0

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

\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

What is 9= 1/(x^2+y^2-7) ?
The solution to 9= 1/(x^2+y^2-7) is Ellipse with (h,k)=(0,0),a= 8/3 ,b= 8/3