Chapter 9.1, Problem 87E

To determine

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

Answer to Problem 87E

The tangent line at a point on a parabola is $y=4x+2$

Explanation of Solution

Given information:

The equation is

  $y=-2x^2,\left(-1,-2\right)$

Calculation:

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

  $\begin{array}{l}y=-2x^2\\ x^2=-\frac{1}{2}y\end{array}$

First find P and Focus

  $\begin{array}{l}4p=-\frac{1}{2}\\ p=-\frac{1}{8}\\ p=-\frac{1}{8}\end{array}$

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

  $\begin{array}{l}\text{Focus = }\left(h,k+p\right)\\ \left(h,k+p\right)=\left(0,0-\frac{1}{8}\right)\\ \left(h,k+p\right)=\left(0,-\frac{1}{8}\right)\end{array}$

So, the slope of the tangent line is

  $y=-2x^2$

Differentiate with respect to x

  $\begin{array}{l}y=-2x^2\\ \frac{dy}{dx}=-4x\\ \frac{dy}{dx}=-4\left(-1\right)\left[\text{since,}x=-1\right]\\ \frac{dy}{dx}=4\end{array}$

  $$ So, the tangent line is

  $\left(y-y_1\right)=m\left(x-x_1\right)$

Put the values of $x_1=-1,y_1=-2\text{ and }m=-4$ on the above equation

  $\begin{array}{l}\left(y-\left(-2\right)\right)=4\left(x-\left(-1\right)\right)\\ \left(y+2\right)=4\left(x+1\right)\\ y+2=4x+4\\ y=4x+4-2\\ y=4x+2\end{array}$

The graph of the parabola and its tangent passing through point $\left(-1,-2\right)$ is shown below:

  

Hence,

The tangent line at a point on a parabola is $y=4x+2$