What is 4(x-2)^2+(y-3)^2-1=0 ?

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

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4(x-2)^2+(y-3)^2-1=0

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

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

Solution steps

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

Rewrite

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

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}

\frac{( ( x - 2 ) ^{2} )}{4} + \frac{( ( y - 4 ) ^{2} )}{16} = 1

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

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

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 2 ) ^{2}}{36} = 1

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