What is 4(x-2)^2+(y-3)^2=1 ?

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

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4(x-2)^2+(y-3)^2=1

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

(h, k) = (2, 3), a = \frac{1}{2}, b = 1

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, 3), a = \frac{1}{2}, b = 1

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, 3), b = 1, a = \frac{1}{2}

4 x^{2} + y^{2} - 16 x + 8 y - 20 = 0

x^{2} - 4 x + 16 y^{2} + 32 y + 4 = 0

9 x^{2} + 12 y^{2} = 108

x^{2} + 3 y^{2} + 4 x + 18 y + 22 = 0

36 x^{2} - 36 x + 64 y^{2} + 256 y + 265 = 576

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