Chapter 8.1, Problem 23E

Solution

To calculate : The equation in the standard form for the parabola that satisfies conditions focus $\left(3,4\right)$ and directrix $y=1$ .

The equation in the standard form for the parabola that satisfies conditions focus $\left(3,4\right)$ and directrix $y=1$ is ${\left(x-3\right)}^2=6\left(y-\frac{5}{2}\right)$ .

Given information :

The parabola with focus $\left(3,4\right)$ and directrix $y=1$ .

Formula used :

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

Calculation :

Consider the parabola with focus $\left(3,4\right)$ and directrix $y=1$ ,

The directrix is said to be the horizontal and is located below the focus, so the parabola opens downward.

The directrix is $y=k-p$ and focus is at $\left(3,4\right)$ .

Equate $k+p$ to $4$ and $h$ to $3$ .

  $k+p=4$ .….. (1)

The directrix is $y=1$ , so,

  $\begin{array}{c}k-p=1\\ k=1+p\end{array}$   .....,. (2)

Substitute the value of (2) into (1).

  $\begin{array}{c}1+p+p=4\\ 1+2p=4\\ 2p=3\\ p=\frac{3}{2}\end{array}$

Substitute $p=\frac{3}{2}$ into the equation (2).

  $\begin{array}{c}k=1+\frac{3}{2}\\ =\frac{2+3}{2}\\ =\frac{5}{2}\end{array}$

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

  $\begin{array}{c}{\left(x-3\right)}^2=4\left(\frac{3}{2}\right)\left(y-\frac{5}{2}\right)\\ {\left(x-3\right)}^2=6\left(y-\frac{5}{2}\right)\end{array}$

Thus, the equation in the standard form for the parabola that satisfies conditions focus $\left(3,4\right)$ and directrix $y=1$ is ${\left(x-3\right)}^2=6\left(y-\frac{5}{2}\right)$ .