What is (x-4)^2+y^2= 1/4 ((x+2)^2+(y-y)) ?

The solution to (x-4)^2+y^2= 1/4 ((x+2)^2+(y-y)) is Ellipse with (h,k)=(6,0),a=4,b=2sqrt(3)

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(x-4)^2+y^2= 1/4 ((x+2)^2+(y-y))

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

(h, k) = (6, 0), a = 4, b = 23

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 6 ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{( 2 3 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (6, 0), a = 4, b = 23

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (6, 0), a = 4, b = 23

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

x^{2} + 4 y^{2} + 10 x + 8 y - 13 = 0

25 x^{2} + 4 y^{2} - 100 x + 8 y + 91 = 0

4 x^{2} + 9 y^{2} - 54 y + 45 = 0

x^{2} + 4 y^{2} = 5

What is (x-4)^2+y^2= 1/4 ((x+2)^2+(y-y)) ?
The solution to (x-4)^2+y^2= 1/4 ((x+2)^2+(y-y)) is Ellipse with (h,k)=(6,0),a=4,b=2sqrt(3)