Chapter 2, Problem 11P

To determine

To find:The point $P$ and $Q$ on the parabola $y=1-x^2$ .

Answer to Problem 11P

The value of the points $P$ and $Q$ on the parabola $y=1-x^2$ are $P=\left(\frac{\sqrt{3}}{2},\frac{1}{4}\right),Q=\left(-\frac{\sqrt{3}}{2},\frac{1}{4}\right)$ .

Explanation of Solution

Given information:

The given equation for parabola is $y=1-x^2$ .

The given equilateral triangle is shown in figure (1).

  

Figure (1)

Calculation:

The parabola is $y=1-x^2$ .

The triangle $ABC$ is an equilateral shown in figure (2).

  

Figure (2)

Each angle in triangle $ABC$ is equal to $60°$ , the slope of the tangent $AB$ which makes an angle of $60°$ with positive x-axis.

  $\begin{array}{c}{m}_1=\tan 60°\\ =\sqrt{3}\end{array}$

The slope of the $AC$ which makes an angle of $120°$ with positive x-axis.

  $\begin{array}{c}{m}_2=\tan 120°\\ =\tan \left(180°-60°\right)\\ =-\tan 60°\\ =-\sqrt{3}\end{array}$

The slope of the tangent to the parabola $y=1-x^2$ at $P$ , $Q$ are as.

  $y\text{'}=-2x$

Substitute $\sqrt{3}$ for $m_1$ and, $-\sqrt{3}$ for $m_2$ in above equation.

  $\begin{array}{l}\pm \sqrt{3}=-2x\\ x=\pm \frac{\sqrt{3}}{2}\end{array}$

Calculate the values of $P$ , $Q$ points on parabola.

Substitute $\pm \frac{\sqrt{3}}{2}$ for $x$ in equation $y=1-x^2$ .

  $\begin{array}{c}y=1-{\left(\pm \frac{\sqrt{3}}{2}\right)}^2\\ =1-\frac{3}{4}\\ =\frac{1}{4}\end{array}$

The points $P$ and $Q$ are $P=\left(\frac{\sqrt{3}}{2},\frac{1}{4}\right),Q=\left(-\frac{\sqrt{3}}{2},\frac{1}{4}\right)$

Therefore, the value of the points $P$ and $Q$ on the parabola $y=1-x^2$ are $P=\left(\frac{\sqrt{3}}{2},\frac{1}{4}\right),Q=\left(-\frac{\sqrt{3}}{2},\frac{1}{4}\right)$ .