What is (3x-3)^2+(2y-4)^2=43+9+16 ?

The solution to (3x-3)^2+(2y-4)^2=43+9+16 is Ellipse with (h,k)=(1,2),a=(2sqrt(17))/3 ,b=sqrt(17)

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(3x-3)^2+(2y-4)^2=43+9+16

(3 x - 3)^{2} + (2 y - 4)^{2} = 43 + 9 + 16

(h, k) = (1, 2), a = \frac{2 17}{3}, b = 17

Solution steps

(3 x - 3)^{2} + (2 y - 4)^{2} = 43 + 9 + 16

Rewrite

(3 x - 3)^{2} + (2 y - 4)^{2} = 43 + 9 + 16

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{( \frac{2 17}{3} ) ^{2}} + \frac{( y - 2 ) ^{2}}{( 17 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, 2), a = \frac{2 17}{3}, b = 17

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, 2), b = 17, a = \frac{2 17}{3}

x^{2} + y^{2} + 12 y = 0

9 x^{2} + 25 y^{2} + 108 x - 200 y + 4 = 0

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

\frac{x ^{2}}{16} + \frac{( y - 1 ) ^{2}}{36} = 1

\frac{2 x - x ^{2} - y ^{2}}{x ^{2} + y ^{2} - x} \leq 0

What is (3x-3)^2+(2y-4)^2=43+9+16 ?
The solution to (3x-3)^2+(2y-4)^2=43+9+16 is Ellipse with (h,k)=(1,2),a=(2sqrt(17))/3 ,b=sqrt(17)