Chapter 12, Problem 8PS

To determine

To graph: The two given parabolas and to find equations of the two lines that are each simultaneously tangent to both parabolas.

Answer to Problem 8PS

The required tangent lines are $y=-2x-1andy=4x-4$ .

Explanation of Solution

Given Information: Given parabolas are $y=x^{2}andy=-x^2+2x-5$ .

Graphs:

  

The red colored graph is of $y=x^2$ and the blue colored graph is of $y=-x^2+2x-5$ .

Calculation:

The two tangent lines are coincident if they have the same slope and same $y-\mathrm{int}ercept.$

For $y=x^2$ , let the point of tangency be $(t,t^2)$ .

Slope, $m_1=\underset{h\to 0}{\lim }\frac{{(x+h)}^2-x^2}{h}=\underset{h\to 0}{\lim }\frac{x^2+2xh+h^2-x^2}{h}\underset{h\to 0}{\lim }\frac{2xh+h^2}{h}\underset{h\to 0}{\lim }\frac{h(2x+h)}{\begin{array}{l}h\\ \end{array}}$

  $=\underset{h\to 0}{\lim }(2x+h)=2x+0=2x$

  $\Rightarrow m_1=2x$ $-(i)$

  $\therefore$ Tangent line is $y_1-t=2x(x-t)$

  $\Rightarrow y_1=2tx-t^2$ $-(ii)$

For $y=-x^2+2x-5$ , let the point of tangency be $(t,-t^2+2t-5)$ .

Slope, $m_2=\underset{h\to 0}{\lim }\frac{[-{(x+h)}^2+2(x+h)-5]-(-x^2+2x-5)}{h}$

  $=\underset{h\to 0}{\lim }\frac{(-x^2-2hx-h^2+2x+2h-5)-(-x^2+2x-5)}{h}$

  $=\underset{h\to 0}{\lim }\frac{-2hx-h^2+2h}{h}$

  $=\underset{h\to 0}{\lim }(-2x-h+2)$

  $=-2x+2$

  $\therefore m_2=-2x+2$ $-(iii)$

  $\therefore$ Tangent line is $y_2-(-t^2+2t-5)=(-2t+2)(x-t)$

  $\Rightarrow y_2+t^2-2t+5=(-2t+2)(x-t)$

  $\Rightarrow y_2=2x-2tx+t^2-5$ $-(iv)$

Equating the slopes, i.e., $m_1=m_2$

  $\Rightarrow 2x=-2x+2$ (Using $(i)and(iii)$ )

  $\begin{array}{l}\Rightarrow 4x=2\\ \Rightarrow x=\frac{1}{2}\end{array}$

Putting the value of $x$ in $(ii)$ ,

  $\begin{array}{l}{y}_1=2t\left(\frac{1}{2}\right)-t^2\\ \Rightarrow y_1=t-t^2\end{array}$

Putting the value of $x$ in $(iv)$ ,

  $\begin{array}{l}{y}_2=2\left(\frac{1}{2}\right)-2t\left(\frac{1}{2}\right)+t^2-5\\ \Rightarrow y_2=t^2-t-4\end{array}$

Equating the $y-\mathrm{int}ercepts$ ,

  $\begin{array}{l}t-t^2=t^2-t-4\\ \Rightarrow 2t^2-2t-4=0\\ \Rightarrow t^2-t-2=0\\ \Rightarrow (t-2)(t+1)=0\\ \Rightarrow t=-1,2\end{array}$

So, the tangent lines are:

  $y=2(-1)x-{(-1)}^2$ (Putting $t=-1$ in $(ii)$ )

  $\Rightarrow y=-2x-1$

and $y=2(2)x-2^2$ (Putting $t=2$ in $(ii)$ )

  $\Rightarrow y=4x-4$

Thus, the required tangent lines are $y=-2x-1andy=4x-4$ .