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

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

100 x^{2} + 4 y^{2} + 1200 x - 32 y + 3264 = 0

16 x^{2} + 15 y^{2} - 64 x - 90 y - 41 = 0

25 y^{2} + 9 x^{2} - 108 x - 200 y + 499 = 0

50 = \frac{100}{1 + x ^{2} + y ^{2}}

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