What is 200=(1000)/(1+x^2+y^2) ?

The solution to 200=(1000)/(1+x^2+y^2) is Ellipse with (h,k)=(0,0),a=2,b=2

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200=(1000)/(1+x^2+y^2)

200 = \frac{1000}{1 + x ^{2} + y ^{2}}

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

Solution steps

200 = \frac{1000}{1 + x ^{2} + y ^{2}}

Rewrite

200 - \frac{1000}{1 + x ^{2} + y ^{2}} = 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

\frac{( x - 8 ) ^{2}}{42} + \frac{( y + 7 ) ^{2}}{16} = 1

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

6 x^{2} + 9 y^{2} + 54 y = 60 x - 87

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

(\frac{x ^{2}}{9}) + (\frac{y ^{2}}{5}) = 1

What is 200=(1000)/(1+x^2+y^2) ?
The solution to 200=(1000)/(1+x^2+y^2) is Ellipse with (h,k)=(0,0),a=2,b=2