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}

\frac{x ^{2}}{8} + \frac{y ^{2}}{a} = 1

2 x^{2} + 4 y^{2} + 6 x - 8 y - 5 = 0

1 - x^{2} - y^{2} = 5

36 (x - \frac{1}{2})^{2} - \frac{1}{4} + 64 (y + 2)^{2} - 4 + 265 = 576

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{6 ^{2}} = 1

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