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

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

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

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

(h, k) = (3, - 3), a = 1, b = 16

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 3), a = 1, b = 16

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 3), b = 16, a = 1

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