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

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

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

4 (x - 3)^{2} + (y - 2)^{2} = 4

(h, k) = (3, 2), a = 1, b = 2

Solution steps

4 (x - 3)^{2} + (y - 2)^{2} = 4

Rewrite

4 (x - 3)^{2} + (y - 2)^{2} = 4

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, 2), a = 1, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 2), b = 2, a = 1

\frac{x ^{2} + y ^{2}}{x} = 16

4 x^{(2)} + 16 y^{(2)} - 16 x - 48 = 0

\frac{( x ^{2} )}{( 16 )} + \frac{( y ^{2} )}{( 12 )} = 1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 7 ) ^{2}}{64} = 1

2 x^{2} + 3 y^{2} + 4 x = 19

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