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

The solution to (3y+3)^2+x=(x-4)^2 is Hyperbola with (h,k)=(9/2 ,-1),a=(sqrt(17))/2 ,b=(sqrt(17))/6

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

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

(h, k) = (\frac{9}{2}, - 1), a = \frac{17}{2}, b = \frac{17}{6}

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( x - \frac{9}{2} ) ^{2}}{( \frac{17}{2} ) ^{2}} - \frac{( y - ( - 1 ) ) ^{2}}{( \frac{17}{6} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (\frac{9}{2}, - 1), a = \frac{17}{2}, b = \frac{17}{6}

4 y^{2} - 9 x^{2} + 16 y + 18 x = 29

- y^{2} + z^{2} = 1

4 y^{2} - x^{2} + 40 y - 4 x + 60 = 0

\frac{( x + 5 ) ^{2}}{4} - \frac{y ^{2}}{4} = 1

9 x^{2} - 4 y^{2} - 18 x - 16 y + 29 = 0

What is (3y+3)^2+x=(x-4)^2 ?
The solution to (3y+3)^2+x=(x-4)^2 is Hyperbola with (h,k)=(9/2 ,-1),a=(sqrt(17))/2 ,b=(sqrt(17))/6