What is 4(x+3)^2+y^2=16 ?

The solution to 4(x+3)^2+y^2=16 is Ellipse with (h,k)=(-3,0),a=2,b=4

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4(x+3)^2+y^2=16

4 (x + 3)^{2} + y^{2} = 16

(h, k) = (- 3, 0), a = 2, b = 4

Solution steps

4 (x + 3)^{2} + y^{2} = 16

Rewrite

4 (x + 3)^{2} + y^{2} = 16

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{2 ^{2}} + \frac{( y - 0 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, 0), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, 0), b = 4, a = 2

2 (1 + 2 y) (y - 2) = - (2 x - 3) (x - 7)

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{36} = 1

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

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

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

What is 4(x+3)^2+y^2=16 ?
The solution to 4(x+3)^2+y^2=16 is Ellipse with (h,k)=(-3,0),a=2,b=4