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

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

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

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

(h, k) = (2, - 3), a = 1, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 3), a = 1, 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 = 1

81 x^{2} + 64 y^{2} = 25

16 x^{2} + 9 y^{2} = 1

7 x^{2} + 16 y^{2} - 28 x + 128 y + 172 = 0

x^{2} - 16 x + 4 y^{2} - 12 y + 16 = 0

5 x^{2} + 3 y^{2} - 15 = 0

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