Chapter 12, Problem 8CR

To determine

To find: The equation of a parabola.

Answer to Problem 8CR

The equation of the parabola is $x^2+2x-4y+9=0$ .

Explanation of Solution

Given information:

The given focus at $\left(-1,3\right)$ and a vertex at $\left(-1,2\right)$ .

Calculation:

The general equation of parabola is,

  ${\left(x-x_1\right)}^2=4a\left(y-y_1\right)\mathrm{.......................}\left(1\right)$

The distance of vertex from direction is equal to the distance of vertex from focus.

Therefore,

  $\begin{array}{c}a=\sqrt{{\left[\left(-1\right)-\left(-1\right)\right]}^2+{\left[3-2\right]}^2}\\ =\sqrt{0^2+1^2}\\ =\sqrt{1}\\ =1\end{array}$

Substitute the value of $\left(x_1,y_1\right)$ and $a$ in equation $\left(1\right)$ .

  $\begin{array}{c}{\left(x-x_1\right)}^2=4a\left(y-y_1\right)\\ {\left[x-\left(-1\right)\right]}^2=4\times 1\times \left[y-2\right]\\ {\left(x+1\right)}^2=4\left(y-2\right)\\ x^2+2x-4y+9=0\end{array}$

Therefore, the equation of the parabola is $x^2+2x-4y+9=0$ .