Chapter 4.2, Problem 20E

a.

To determine

To calculate:The equation of line that is tangent to the given curve at the given point.

Answer to Problem 20E

The equation of tangent is $y+x+1=0$

Explanation of Solution

Given information:The equation of the curve is $y^2-2x-4y-1=0$ and the point is $\left(x_1,y_1\right)=\left(-2,1\right)$ .

Formula used:

Slope-point form of line is

  $\left(y-y_1\right)=m\left(x-x_1\right),\text{where}\left(x_1,y_1\right)\text{is}\text{the}\text{intercept}\text{and}m\text{is}\text{the}\text{slope}\text{of}\text{the}\text{line}$.

Also product, quotient and chain rule of derivatives are used.

Calculation:

The curve equation is $y^2-2x-4y-1=0$ ................(1)

And the point at which the tangent equation is to be determined is $\left(x_1,y_1\right)=\left(-2,1\right)$ .

By differentiating equation (1) we get

  $\begin{array}{l}{y}^2-2x-4y-1=0\\ \frac{d{y}^2}{dx}-2\frac{dx}{dx}-4\frac{dy}{dx}-0=0\\ 2y\frac{dy}{dx}-2-4\frac{dy}{dx}=0\mathrm{........}\left(2\right)\\ \text{Now}\text{put}\text{the}\text{given}\text{point}\left(x_1,y_1\right)=\left(-2,1\right)\text{in}\text{equation}\left(2\right)\\ 2\left(1\right)\frac{dy}{dx}-2-4\frac{dy}{dx}=0\\ \frac{dy}{dx}\left(2-4\right)-2=0\\ \frac{dy}{dx}\left(-2\right)=2\\ \frac{dy}{dx}=m=-1\end{array}$

Here $\frac{dy}{dx}=m=-1$ is the slope of the tangent.

The slope point form of the tangent line is

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}\left(x_1,y_1\right)=\left(-2,1\right)\text{and}m=-1\\ \left(y-1\right)=-1\left(x-\left(-2\right)\right)\\ \left(y-1\right)=-1\left(x+2\right)\\ y-1=-x-2\\ y+x+1=0\end{array}$

Hence the equation of tangent is $y+x+1=0$ .

b.

To determine

To calculate:The equation of line that is normal to the given curve at the given point.

Answer to Problem 20E

The equation of normal to the curve is $y-x-3=0$

Explanation of Solution

Given information:The equation of the curve is $y^2-2x-4y-1=0$ and the point is $\left(x_1,y_1\right)=\left(-2,1\right)$ .

Formula used:

The slope of normal to the curve at a given point is

  $m_1=-\frac{dx}{dy}=-\frac{1}{m}$

Where m is the slope of the tangent to the curve at that point.

Calculation:

  $\begin{array}{l}\text{Since}m=-1\text{then}\text{the}\text{slope}\text{of}\text{normal}\text{is}{m}_1\text{=}-\frac{1}{m}=1\end{array}$

Then the equation of normal at $\left(x_1,y_1\right)=\left(-2,1\right)$ is

  $\begin{array}{l}\left(y-y_1\right)=m_1\left(x-x_1\right)\\ \left(y-1\right)=1\left(x-\left(-2\right)\right)\\ y-1=x+2\\ y-x-3=0\end{array}$

Hence the equation of normal to the curve is $y-x-3=0$ .