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

The solution to ((x+1)/2)^2+((y-3)/4)^2=4 is Ellipse with (h,k)=(-1,3),a=4,b=8

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

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

(h, k) = (- 1, 3), a = 4, b = 8

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{4 ^{2}} + \frac{( y - 3 ) ^{2}}{8 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 3), a = 4, b = 8

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 3), b = 8, a = 4

\frac{( x - 3 ) ^{2}}{( 18 )} + \frac{( y + 1 ) ^{2}}{12} = 1

4 x^{2} + 8 y^{2} - 16 \cdot + 32 y + 32 = 0

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

y^{2} = \frac{4 - x ^{2}}{5}

\frac{( x + 4 ) ^{2}}{64} + \frac{( y + 2 ) ^{2}}{100} = 1

What is ((x+1)/2)^2+((y-3)/4)^2=4 ?
The solution to ((x+1)/2)^2+((y-3)/4)^2=4 is Ellipse with (h,k)=(-1,3),a=4,b=8