Chapter 8.1, Problem 24E

Solution

To calculate : The equation in the standard form for the parabola that satisfies conditions focus $\left(2,-3\right)$ and directrix $x=5$ .

The equation in the standard form for the parabola that satisfies conditions focus $\left(2,-3\right)$ and directrix $x=5$ is ${\left(y+3\right)}^2=-6\left(x-\frac{7}{2}\right)$ .

Given information :

The parabola with focus $\left(2,-3\right)$ and directrix $x=5$ .

Formula used :

The equation for parabola for which vertex is at $\left(h,k\right)$ and focus is at $\left(h-p,k\right)$ is ${\left(y-k\right)}^2=4p\left(x-h\right)$ .

Calculation :

Consider the parabola with focus $\left(2,-3\right)$ and directrix $x=5$ ,

The parabola opens to the left and the vertex is situated halfway between focus and directrix on $y=k$ axis.

This implies that $k=-3$ .

Also,

  $\begin{array}{c}h=\frac{2+5}{2}\\ =\frac{7}{2}\end{array}$

Equate $h-p$ to $5$ and substitute $-4$ for $h$ .

  $\begin{array}{c}h-p=5\\ \frac{7}{2}-p=5\\ p=-\frac{3}{2}\end{array}$

Substitute the desired values to the equation ${\left(y-k\right)}^2=4p\left(x-h\right)$ .

  $\begin{array}{c}{\left(y-\left(-3\right)\right)}^2=4\left(\frac{-3}{2}\right)\left(x-\frac{7}{2}\right)\\ {\left(y+3\right)}^2=-6\left(x-\frac{7}{2}\right)\end{array}$

Thus, the equation in the standard form for the parabola that satisfies conditions focus $\left(2,-3\right)$ and directrix $x=5$ is ${\left(y+3\right)}^2=-6\left(x-\frac{7}{2}\right)$ .