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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

(x^{2} + 10 x) + (4 y^{2} + 9 y) = 7

8 x^{2} - 176 x + 9 y^{2} - 18 y + 905 = 0

9 x^{2} + 25 y^{2} = 400

36 - 4 x^{2} - 9 y^{2} = 0

\frac{12 ( x - 1 ) ^{2}}{24} + \frac{3 y ^{2}}{24} = \frac{24}{24}

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