What is ((x-5)^2)/4+((y+3)^2)/(16)=1 ?

The solution to ((x-5)^2)/4+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(5,-3),a=2,b=4

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((x-5)^2)/4+((y+3)^2)/(16)=1

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

(h, k) = (5, - 3), a = 2, b = 4

Solution steps

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{2 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, - 3), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, - 3), b = 4, a = 2

x^{2} = 6 y

\frac{( y - 3 ) ^{2}}{11} - \frac{( x + 1 ) ^{2}}{5} = 1

x^{2} + 2 x + y^{2} + 4 y = 20

x - y^{2} = 0

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

What is ((x-5)^2)/4+((y+3)^2)/(16)=1 ?
The solution to ((x-5)^2)/4+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(5,-3),a=2,b=4