What is ((x-3)^2+y^2)/((x+3)^2+y^2)>= 4 ?

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

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((x-3)^2+y^2)/((x+3)^2+y^2)>= 4

\frac{( x - 3 ) ^{2} + y ^{2}}{( x + 3 ) ^{2} + y ^{2}} \geq 4

(h, k) = (- 5, 0), a = 4, b = 4

Solution steps

\frac{( x - 3 ) ^{2} + y ^{2}}{( x + 3 ) ^{2} + y ^{2}} \geq 4

Rewrite

\frac{( x - 3 ) ^{2} + y ^{2}}{( x + 3 ) ^{2} + y ^{2}} \geq 4

in the form of the standard ellipse equation

\frac{( x - ( - 5 ) ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 5, 0), a = 4, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 5, 0), b = 4, a = 4

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

4 x^{2} + 9 y^{2} - 8 x = 32

x^{2} + 4 y^{2} + 4 y = 8

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

4 x^{2} + 9 y^{2} - 8 x + 18 y - 23 = 0

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