What is ((x+2)^2}{16}+\frac{(y-5)^2)/4 =1 ?

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

x^{2} + 9 y^{2} - 14 x + 36 y + 49 = 0

13 x^{2} - 10 x + 2 + y^{2} = 4 x - 1

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

- 3 y^{(2)} - 2 x^{(2)} = 2

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