What is 8x^2+3y-12x+6y+18=0 ?

The solution to 8x^2+3y-12x+6y+18=0 is Parabola with (h,k)=(3/4 ,-3/2),p=-9/32

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8x^2+3y-12x+6y+18=0

8 x^{2} + 3 y - 12 x + 6 y + 18 = 0

(h, k) = (\frac{3}{4}, - \frac{3}{2}), p = - \frac{9}{32}

Solution steps

8 x^{2} + 3 y - 12 x + 6 y + 18 = 0

Rewrite

8 x^{2} + 3 y - 12 x + 6 y + 18 = 0

in the standard form

4 (- \frac{9}{32}) (y - (- \frac{3}{2})) = (x - \frac{3}{4})^{2}

Therefore parabola properties are:

(h, k) = (\frac{3}{4}, - \frac{3}{2}), p = - \frac{9}{32}

x^{2} + 6 x + 4 y + 8 = 0

4 x^{2} = 128 y

(y - 4)^{2} = 32 (x - 6)

5 - \frac{( 4 y )}{3} = \frac{( x ^{2} )}{9}

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

What is 8x^2+3y-12x+6y+18=0 ?
The solution to 8x^2+3y-12x+6y+18=0 is Parabola with (h,k)=(3/4 ,-3/2),p=-9/32