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

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

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

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

(h, k) = (3, - 4), a = 2, b = 22

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 4), a = 2, b = 22

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 4), b = 22, a = 2

3 x^{2} - 12 x + y^{2} - 4 y + 1 = 0

(\frac{u + v}{u ^{2} + v ^{2}}) + (\frac{- v + u}{u ^{2} + v ^{2}}) = 1

3 x^{2} + 2 y^{2} - 6 = 0

\frac{y ^{2}}{64} + \frac{x ^{2}}{100} = 1

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{b} = 1

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