Chapter 10.8, Problem 41E

To determine

Find the value of $k$ so that the graphs are tangents to each other.

Answer to Problem 41E

  $k=-\frac{9}{8}$

Explanation of Solution

Given information:

  $x=2y^{2}andx+3y=k$

Calculation:

Here, we will consider:

  $\begin{array}{l}x=2y^2\mathrm{...........}\left(1\right)\\ x+3y=k\mathrm{.........}\left(2\right)\end{array}$

Now, we will replace $x=2y^{2}inequation\left(2\right)$ :

  $\begin{array}{l}x+3y=k\\ \left(2y^2\right)+3y=k\\ 2y^2+3y-k=0\\ y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ y=\frac{-3\pm \sqrt{{\left(3\right)}^2-4\left(2\right)\left(-k\right)}}{2a}\end{array}$

Now, to become a tangent the line has to touch only in one point.

Thus, the discriminator has to be zero for only one solution.

  $\begin{array}{l}{b}^2-4ac=0\\ {\left(3\right)}^2-4\left(2\right)\left(-k\right)=0\\ 9+8k=0\\ k=-\frac{9}{8}\end{array}$

Hence, the required answer is $k=-\frac{9}{8}$ .