Chapter 8.1, Problem 25E

Solution

To calculate : The equation in the standard form for the parabola that satisfies conditions vertex $\left(4,3\right)$ and directrix $x=6$ .

The equation in the standard form for the parabola that satisfies conditions vertex $\left(4,3\right)$ and directrix $x=6$ is ${\left(y-3\right)}^2=-8\left(x-4\right)$ .

Given information :

The parabola with vertex $\left(4,3\right)$ and directrix $x=6$ .

Formula used :

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

Calculation :

Consider the parabola with vertex $\left(4,3\right)$ and directrix $x=6$ ,

Equate $h-p$ to $6$ and substitute $4$ for $h$ .

  $\begin{array}{c}h-p=6\\ 4-p=6\\ p=-2\end{array}$

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

  $\begin{array}{c}{\left(y-3\right)}^2=4\left(-2\right)\left(x-4\right)\\ {\left(y-3\right)}^2=-8\left(x-4\right)\end{array}$

Thus, the equation in the standard form for the parabola that satisfies conditions vertex $\left(4,3\right)$ and directrix $x=6$ is ${\left(y-3\right)}^2=-8\left(x-4\right)$ .