Chapter 11.1, Problem 50E

To determine

To find: the equation of the parabola.

Answer to Problem 50E

The equation of parabola is $x^2=-2\left(-1+\sqrt{5}\right)$ .

Explanation of Solution

The equation of parabola is $x^2=-2\left(-1+\sqrt{5}\right)$ .

Given information:

From the given figure it can be observed that,

The graph of the parabola opens to the upwards and parabola has vertical axis.

Slope of the line which going through the focus of parabola.

The sloping line x-coordinate is $2$ .

Calculation:here, the graph of parabola opens to the downwards, so $p>0$ .

Because the parabola has vertical axis,

So the equation of parabola should be in the following form,

  $x^2-4py$

The line equation is $y=mx+c$ .

Line passes form the focus so, substitute the value of $x$ by $0$ and $y$ by $p$ ,

So,

  $\begin{array}{l}y=mx+c\\ p=m·0+c\text{ [substitute }x\text{ by 0 and }y\text{ by p]}\\ p=c\end{array}$

Since $m=\frac{1}{2},c=p,x=2$ then,

  $\begin{array}{l}y=mx+c\\ y=\frac{1}{2}\left(2\right)+p\text{ [substitute m by}\frac{1}{2},c\text{ by p and }x\text{ by 2]}\\ y=1+p\end{array}$

So the point on parabola will be $\left(2,1+p\right)$ , so the point satisfy the equation of parabola,

  $\begin{array}{l}{x}^2-4py=0\\ {\left(2\right)}^2=4p\left(1+p\right)\text{ [substitute }x\text{ by 2 and }y\text{ by 1+p]}\\ 4=4p+4p^2\text{ [multiply]}\\ p^2+p-1=0\text{ [divide by 4 and subtract 1]}\\ p=\frac{-1\pm \sqrt{{\left(1\right)}^2-4·1·\left(-1\right)}}{2·1}\\ p=\frac{-1\pm \sqrt{5}}{2}\text{ [simplify]}\end{array}$

So the equation of parabola is,

  $\begin{array}{l}{x}^2-4py\\ x^2-4\left(\frac{-1+\sqrt{5}}{2}\right)y\text{ [substitute p by}\frac{-1+\sqrt{5}}{2}\text{ ]}\\ x^2+2\left(-1+\sqrt{5}\right)y\text{ [divide ]}\\ x^2+2\left(-1+\sqrt{5}\right)=0\text{ [simplify]}\\ x^2=-2\left(-1+\sqrt{5}\right)\end{array}$

Hence, the equation of parabola is $x^2=-2\left(-1+\sqrt{5}\right)$ .