Chapter 8, Problem 39RE

Solution

To find: the equation for the conic in the standard form.

The standard equation of the parabola opens upwards with vertex $\left(-3,3\right)$ is $x^2+6x-12y+45=0$ .

Given data:

Type of conic: Parabola

Vertex of the parabola $=\left(-3,3\right)$

Equation of the directrix: $y=0$

Formula used:

A parabola is the set of all points in a plane equidistant from a particular line called directrix and a particular point called focus in a plane.

The standard equation of the parabola with vertex $\left(h,k\right)$ opens upwards or downwards is given as:

  ${\left(x-h\right)}^2=4p\left(y-k\right)$

Calculation:

It is given that vertex of the parabola $=\left(-3,3\right)$ and equation of the directrix is $y=0$ . So, it can be observed that the parabola opens upwards.

  $h=-3,k=3$

Calculate $p$ . It is given that $y=0$ is the equation of directrix of the parabola.

  $\begin{array}{c}y=0\\ k-p=0\left(y=k-p\right)\\ 3-p=0\\ p=3\end{array}$

The standard equation of the parabola with vertex $\left(-3,3\right)$ opens upwards is given as:

  $\begin{array}{c}{\left(x-\left(-3\right)\right)}^2=4\times 3\left(y-3\right)\\ {\left(x+3\right)}^2=12\left(y-3\right)\\ x^2+6x+9=12y-36\\ x^2+6x-12y+45=0\end{array}$

Therefore, the standard equation of the parabola opens upwards with vertex $\left(-3,3\right)$ is $x^2+6x-12y+45=0$ .