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

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

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

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

(h, k) = (2, - 3), a = 2, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 3), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, - 3), b = 4, a = 2

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

\frac{x ^{2}}{5.6 6 ^{2}} + \frac{y ^{2}}{3.5 9 ^{2}} = 1

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

5 x^{2} + 2 y^{2} = 2

3 (x - 5)^{2} + 4 (y + 1)^{2} = 6

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