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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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