Answer :
Answer:
The standard form of the parabola is [tex](x-3)^2=4*\frac{3}{8}(y+2)[/tex]
Step-by-step explanation:
The standard form of a parabola is
[tex](x-h)^2=4p(y-k)[/tex].
In order to convert [tex]2x^2-12x-3y+12=0[/tex] into the standard form, we first separate the variables:
[tex]2x^2-12x+3y+12=0\\\\2x^2-12x+12=3y[/tex]
we now divided both sides by 2 to remove the coefficient from [tex]2x^2[/tex] and get:
[tex]x^2-6x+6=\frac{3}{2}y[/tex].
We complete the square on the left side by adding 3 to both sides:
[tex]x^2-6x+6+3=\frac{3}{2}y+3[/tex]
[tex]x^2-6x+9=\frac{3}{2}y+3[/tex]
[tex](x-3)^2=\frac{3}{2}y+3[/tex]
now we bring the right side into the form [tex]4p(y-k)[/tex] by first multiplying the equation by [tex]\frac{2}{3}[/tex]:
[tex]\frac{2}{3} *(x-3)^2=\frac{2}{3} *(\frac{3}{2}y+3)\\\\\frac{2}{3} *(x-3)^2=y+2[/tex]
and then we multiplying both sides by [tex]\frac{3}{2}[/tex] to get
[tex](x-3)^2=\frac{3}{2} (y+2)[/tex].
Here we see that
[tex]4p=\frac{3}{2}[/tex]
[tex]\therefore p=\frac{3}{8}[/tex]
Thus, finally we have the equation of the parabola in the standard form:
[tex]\boxed{(x-3)^2=4*\frac{3}{8}(y+2)}[/tex]