Chapter 12.7, Problem 47E

To determine

To find: the equation of the parabola in general form that has horizontal axis and passes through the given points.

Answer to Problem 47E

  $y^2+3x+7y=0.$

Explanation of Solution

Given:

The points are:

  $\left(0,0\right),\left(2,-1\right)\text{ and }\left(4,-4\right)$ .

Concept used:

The parabola equation:

  $x^2+y+Dx+Ey+F=0.$

Where $\left(x,y\right)$ are the coordinates.

Calculation:

Here the parabola has horizontal axis. Its directrix is parallel to $y$ -axis.

The equation of a parabola when the directrix is parallel to $y$ -axis is in the form:

  $y^2+Dx+Ey+F=\mathrm{0.....}\left(i\right)$ .

Since, the equation $\left(i\right)$ is passing through

  $\left(0,0\right),\left(2,-1\right)\text{ and }\left(4,-4\right)$ .

So, all these points satisfy the equation $\left(i\right)$

Now substitute $\left(0,0\right)$ in the equation $\left(i\right)$ to obtain as:

  $\begin{array}{l}{\left(0\right)}^2+D\left(0\right)+E\left(0\right)+F=0.\\ F=0.\end{array}$

So, the equation $\left(i\right)$ becomes:

  $y^2+Dx+Ey=\mathrm{0.....}\left(ii\right)$

Now substituting the $\left(2,-1\right)$ in equation $\left(ii\right)$ to obtain:

  $\begin{array}{l}{\left(-1\right)}^2+D\left(2\right)+E\left(-1\right)=0.\\ 1+2D-E=0.\\ E=1+2D\mathrm{.....}\left(iii\right)\end{array}$

Also substituting $\left(4,-4\right)$ in equation $\left(ii\right)$ ,to obtain:

  $\begin{array}{l}{\left(-4\right)}^2+D\left(4\right)+E\left(-4\right)=0.\\ 16+4D-4E=0.\\ 4E=16+4D.\\ E=4+D\mathrm{..........}\left(iv\right)\end{array}$

From the equation $\left(iii\right)$ and $\left(iv\right)$ to obtain:

  $\begin{array}{l}1+2D=4+D.\\ 2D-D=4-1.\\ D=3.\end{array}$

Substitute $D=3$ in equation $\left(iv\right)$ ,then:

  $E=4+3=7.$

Therefore, the required equation of parabola is:

  $y^2+3x+7y=0.$

Hence, $y^2+3x+7y=0.$