Chapter 7.5, Problem 32WE

To determine

The value of $a\text{ and }k$ for which the parabola is $y-k=a{\left(x+3\right)}^2$ contains the points $\left(-5,1\right)\text{ and }\left(1,7\right)$ .

Answer to Problem 32WE

The value of $a=\frac{1}{2}$ and $k=-1$ .

Explanation of Solution

Given information: The equation of the given parabola is,

  $y-k=a{\left(x+3\right)}^2\mathrm{....}\left(1\right)$

Calculation: Since the parabola (1) contains the points $\left(-5,1\right)\text{ and }\left(1,7\right)$ , therefore (1) will satisfy the points $\left(-5,1\right)\text{ and }\left(1,7\right)$ . Therefore from (1),

  $\begin{array}{c}\left(1\right)-k=a{\left(\left(-5\right)+3\right)}^2\mathrm{....}\left(\left(x,y\right)\to \left(-5,1\right)\right)\\ 1-k=4a\\ \text{or, }k=1-4a\mathrm{....}\left(2\right)\end{array}$

Similarly satisfy the point $\left(1,7\right)$ in (1) gives,

  $\begin{array}{c}\left(7\right)-k=a{\left(\left(1\right)+3\right)}^2\mathrm{....}\left(\left(x,y\right)\to \left(1,7\right)\right)\\ 7-k=16a\\ 7-\left(1-4a\right)=16a\mathrm{....}\left(\begin{array}{l}From\left(2\right)\\ k=1-4a\end{array}\right)\\ a=\frac{1}{2}\end{array}$

Now substitute $a=\frac{1}{2}$ in (2) gives $k=-1$ .

Hence $a=\frac{1}{2}$ and $k=-1$ .