Chapter 9, Problem 68SGR

To determine

To find: The where did the support beams meet to parabola.

Answer to Problem 68SGR

The beam given by equation $y=-x+10$ meets the entrance at points $\left(0,10\right)$ and $\left(30,-10\right)$ .

The beam given by equation $y=x-10$ meets the entrance at points $\left(20,10\right)$ and $\left(-10,-20\right)$ .

Explanation of Solution

Given information:

Equation of the parabola $y=-\frac{1}{10}{\left(x-10\right)}^2+20$

The construction team put in two support beams with the equations $y=-x+10$ and $y=x-10$ .

Calculation:

The front entrance of a building is parabola in shape with the equation,

  $y=-\frac{1}{10}{\left(x-10\right)}^2+20\mathrm{........}(1)$

The parabolic entrance is supported by two beams The support beams have the following equations,

  $y=-x+10\mathrm{.......}(2)$

And

  $y=x-10\mathrm{........}(3)$

The beam given by equation (2) $y=-x+10$ meets the parabola (1) $y=-\frac{1}{10}{\left(x-10\right)}^2+20$ at the intersection point of two equations.

Substitute $y=-x+10$ from equation (2) in equation (1) $y=-\frac{1}{10}{\left(x-10\right)}^2+20$ and solve for x

  $\begin{array}{l}-x+10=-\frac{1}{10}{\left(x-10\right)}^2+20\\ -10x+100=-x^2+20x-100+200\\ x^2-30x=0\end{array}$

  $\begin{array}{l}x\left(x-30\right)=0\\ x=0,30\end{array}$

Substitute $x=0$ in equation (2) $y=-x+10$ and solve for y

  $\begin{array}{l}y=-0+10\\ y=10\end{array}$

Hence, $\left(0,10\right)$ is a solution of parabola and line (2) $y=-x+10$ .

Substitute $x=30$ in equation (2) $y=-x+10$ and solve for y,

  $\begin{array}{l}y=-30+10\\ y=-20\end{array}$

Hence, $\left(30,-10\right)$ is a solution of parabola and line (2) $y=-x+10$ .

Therefore, beam given by equation (2) $y=-x+10$ meets the entrance at points $\left(0,10\right)$ and $\left(30,-10\right)$ .

The beam given by equation (3) $y=x-10$ meets the parabola (1) $y=-\frac{1}{10}{\left(x-10\right)}^2+20$ at the intersection point of two equations.

Substitute $y=x-10$ from equation (3) in equation (1) $y=-\frac{1}{10}{\left(x-10\right)}^2+20$ and solve for x,

  $\begin{array}{l}x-10=-\frac{1}{10}{\left(x-10\right)}^2+20\\ 10x-100=-x^2+20x-100+200\\ x^2-10x-200=0\end{array}$

Simplify further

  $x=20,-10$

Substitute $x=20$ in equation (3) $y=x-10$ and solve for y

  $\begin{array}{l}y=20-10\\ y=10\end{array}$

Hence $\left(20,10\right)$ is a solution of parabola and line (3) $y=x-10$ .

Therefore, beam given by equation (3) $y=x-10$ meets the entrance at points $\left(20,10\right)$ and $\left(-10,-20\right)$ .