What is ((x+2)/4)^2+((y-3)/(16))^2=1 ?

The solution to ((x+2)/4)^2+((y-3)/(16))^2=1 is Ellipse with (h,k)=(-2,3),a=4,b=16

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((x+2)/4)^2+((y-3)/(16))^2=1

(\frac{x + 2}{4})^{2} + (\frac{y - 3}{16})^{2} = 1

(h, k) = (- 2, 3), a = 4, b = 16

Solution steps

(\frac{x + 2}{4})^{2} + (\frac{y - 3}{16})^{2} = 1

Rewrite

(\frac{x + 2}{4})^{2} + (\frac{y - 3}{16})^{2} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 3), a = 4, b = 16

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 3), b = 16, a = 4

4 x^{2} + 9 y^{2} = 2 x

x^{2} + 4 x + 4 y^{2} = 0

2 x^{2} + 4 y^{2} = 18

2 x^{2} + 4 y^{2} = 16

4 x^{2} + 9 y^{2} = 72

What is ((x+2)/4)^2+((y-3)/(16))^2=1 ?
The solution to ((x+2)/4)^2+((y-3)/(16))^2=1 is Ellipse with (h,k)=(-2,3),a=4,b=16