Chapter 8, Problem 40RE

Solution

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

The standard equation of the parabola opens to the left with vertex $\left(1,-2\right)$ is $y^2+4y-8x+12=0$ .

Given data:

Type of conic: Parabola opens to the left

Vertex of the parabola $=\left(1,-2\right)$

Focal Length $=2$

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.

The standard equation of the parabola with vertex $\left(h,k\right)$ opens to the left or right is given as:

  ${\left(y-k\right)}^2=4p\left(x-h\right)$

Calculation:

It is given that vertex of the parabola is $\left(1,-2\right)$ .

  $h=1,k=-2$

Calculate $p$ . The focal length of a parabola with vertex $\left(h,k\right)$ is $p$ .

  $p=2$

The standard equation of the parabola with vertex $\left(1,-2\right)$ opens to the left is given as:

  $\begin{array}{c}{\left(y-\left(-2\right)\right)}^2=4\times 2\left(x-1\right)\\ {\left(y+2\right)}^2=8\left(x-1\right)\\ y^2+4y+4=8x-8\\ y^2+4y-8x+12=0\end{array}$

Therefore, the standard equation of the parabola opens to the left with vertex $\left(1,-2\right)$ is $y^2+4y-8x+12=0$ .