Chapter 8, Problem 38RE

Solution

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

The equation for the conic in the standard form is $x^2=-12y$ .

Given data:

Vertex of Parabola: $\left(0,0\right)$

Focal width $=12$

Type of conic: Parabola opens downwards.

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.

Calculation:

The parabola opens downward. Thus, the focus of the parabola is in the form of $\left(0,p\right)$ .

Calculate $p$ .

  $\begin{array}{l}\text{Focal-width}=\left|4p\right|\\ 12=\left|4p\right|\\ 4p=\pm 12\\ p=\pm 3\end{array}$

The parabola opens downwards then the value of $p$ must be negative that is $p<0$ .

Hence, the focus of parabola is $\left(0,-3\right)$ and the equation of the directrix for this parabola is $y=3$ .

Let $P\left(x,y\right)$ be any arbitrary point equidistant from focus $\left(0,-3\right)$ and the line $y=3$ . Let $d_F$ and $d_D$ are the distance from point $P$ to focus and the directrix respectively.

Calculate $d_F$ and $d_D$ .

  $\begin{array}{l}{d}_F=\sqrt{{\left(x-0\right)}^2+{\left(y-\left(-3\right)\right)}^2}\\ d_D=\sqrt{{\left(x-x\right)}^2+{\left(y-3\right)}^2}\end{array}$

Equate $d_F$ and $d_D$ , square both sides and solve the equation.

  $\begin{array}{c}{d}_F=d_D\\ \sqrt{{\left(x-0\right)}^2+{\left(y-\left(-3\right)\right)}^2}=\sqrt{{\left(x-x\right)}^2+{\left(y-3\right)}^2}\\ {\left(x-0\right)}^2+{\left(y+3\right)}^2={\left(x-x\right)}^2+{\left(y-3\right)}^2\\ x^2+y^2+6y+9=y^2-6y+9\\ x^2=-12y\end{array}$

Therefore, the equation for the conic in the standard form is $x^2=-12y$ .