What is 12(x+4)^2+3(y-1)^2=48 ?

The solution to 12(x+4)^2+3(y-1)^2=48 is Ellipse with (h,k)=(-4,1),a=2,b=4

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12(x+4)^2+3(y-1)^2=48

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

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

Solution steps

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

Rewrite

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 4, 1), b = 4, a = 2

\frac{( x + 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1 \cdot \frac{1}{9} \cdot \frac{1}{4}

16 x^{2} + 9 y^{2} - 32 x + 18 y - 119 = 0

16 x^{2} + 36 y^{2} + 112 x - 108 y + 133 = 0

2 x^{2} + 11 y^{2} = 11

What is 12(x+4)^2+3(y-1)^2=48 ?
The solution to 12(x+4)^2+3(y-1)^2=48 is Ellipse with (h,k)=(-4,1),a=2,b=4