Chapter 9.1, Problem 86E

To determine

To find the tangent line at a point on a parabola.

Answer to Problem 86E

The tangent line at a point on a parabola is $y=-3x-\frac{9}{2}$

Explanation of Solution

Given information:

The equation is

  $x^2=2y,\left(-3,\frac{9}{2}\right)$

Calculation:

The tangent line at a point on a parabola can be obtained as

  $x^2=2y$

First find P and Focus

  $\begin{array}{l}4p=2\\ p=\frac{2}{4}\\ p=\frac{1}{2}\left[\text{divide numerator and denominator by 2}\right]\end{array}$

  $\text{since vertex is }\left(0,0\right)\text{, so }\left(h,k\right)=\left(0,0\right)$

  $\begin{array}{l}\text{Focus = }\left(h,k+p\right)\\ \left(h,k+p\right)=\left(0,0+\frac{1}{2}\right)\\ \left(h,k+p\right)=\left(0,\frac{1}{2}\right)\end{array}$

Find the Y-intercept $\left(0,b\right)$ of the tangent line by equating the lengths of the two sides of the isosceles triangle

The equation for $D_1$ , the distance between focus and the desired Y-intercept

  $D_1=\frac{1}{2}-b$

Find $D_2$ ,the distance between given point and the focus

  $\begin{array}{l}{\text{D}}_2=\sqrt{{\left(\left(-3\right)-0\right)}^2+{\left(\frac{9}{2}-\frac{1}{2}\right)}^2}\left[\begin{array}{l}\text{since }\left(x,y\right)=\left(-3,\frac{9}{2}\right)\\ \text{focus = }\left(0,\frac{1}{2}\right)\end{array}\right]\\ {\text{D}}_2=\sqrt{{\left(-3\right)}^2+{\left(\frac{9-1}{2}\right)}^2}\\ {\text{D}}_2=\sqrt{{\left(-3\right)}^2+{\left(\frac{8}{2}\right)}^2}\\ {\text{D}}_2=\sqrt{9+\frac{64}{4}}\\ {\text{D}}_2=\sqrt{\frac{36+64}{4}}\\ {\text{D}}_2=\sqrt{\frac{100}{4}}\\ {\text{D}}_2=\frac{10}{2}\end{array}$

Therefore,

  ${\text{D}}_1={\text{D}}_2$

  $\begin{array}{l}{\text{D}}_1={\text{D}}_2\\ \frac{1}{2}-b=\frac{10}{2}\\ b=\frac{1}{2}-\frac{10}{2}\\ b=\frac{1-10}{2}\\ b=-\frac{9}{2}\\ b=-\frac{9}{2}\end{array}$

So, the slope of the tangent line is

  $\begin{array}{l}m=\frac{y-b}{x-0}\\ m=\frac{\left(\frac{9}{2}\right)-\left(-\frac{9}{2}\right)}{\left(-3\right)-0}\left[\left(x,y\right)=\left(-3,\frac{9}{2}\right)\right]\\ m=\frac{\frac{9}{2}+\frac{9}{2}}{-3}\\ m=\frac{\frac{9+9}{2}}{-3}\\ m=\frac{18}{-3\left(2\right)}\\ m=\frac{18}{-6}\\ m=-3\left[\text{divide numerator and denominator by 6}\right]\end{array}$

Therefore,

  $\begin{array}{l}y=mx+b\\ y=\left(-3\right)x+\left(-\frac{9}{2}\right)\\ y=-3x-\frac{9}{2}\end{array}$

The graph of the parabola and its tangent passing through point $\left(-3,\frac{9}{2}\right)$ is shown below:

  

Hence,

The tangent line at a point on a parabola is $y=-3x-\frac{9}{2}$