Question

The general form of an parabola is 2x2−12x−3y+12=0 .

Answer 1

Answer:

The standard form of the parabola is $(x-3)^2=4*\frac{3}{8}(y+2)$

Step-by-step explanation:

The standard form of a parabola is

$(x-h)^2=4p(y-k)$.

In order to convert $2x^2-12x-3y+12=0$ into the standard form, we first separate the variables:

$2x^2-12x+3y+12=0\\\\2x^2-12x+12=3y$

we now divided both sides by 2 to remove the coefficient from $2x^2$ and get:

$x^2-6x+6=\frac{3}{2}y$.

We complete the square on the left side by adding 3 to both sides:

$x^2-6x+6+3=\frac{3}{2}y+3$

$x^2-6x+9=\frac{3}{2}y+3$

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

now we bring the right side into the form $4p(y-k)$ by first multiplying the equation by $\frac{2}{3}$:

$\frac{2}{3} *(x-3)^2=\frac{2}{3} *(\frac{3}{2}y+3)\\\\\frac{2}{3} *(x-3)^2=y+2$

and then we multiplying both sides by $\frac{3}{2}$ to get

$(x-3)^2=\frac{3}{2} (y+2)$.

Here we see that

$4p=\frac{3}{2}$

$\therefore p=\frac{3}{8}$

Thus, finally we have the equation of the parabola in the standard form:

$\boxed{(x-3)^2=4*\frac{3}{8}(y+2)}$