Chapter 4.2, Problem 21E

a.

To determine

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

Answer to Problem 21E

The equation of tangent is $7y+6x+6=0$ .

Explanation of Solution

Given information:The equation of the curve is $6x^2+3xy+2y^2+17y-6=0$ and the point is $\left(x_1,y_1\right)=\left(-1,0\right)$ .

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 $6x^2+3xy+2y^2+17y-6=0$.........(1)

And the point at which the tangent equation is to be determined is $\left(x_1,y_1\right)=\left(-1,0\right)$ .

By differentiating equation (1) we get

  $\begin{array}{l}6x^2+3xy+2y^2+17y-6=0\\ 6\frac{d{x}^2}{dx}+3\frac{d\left(xy\right)}{dx}+2\frac{d{y}^2}{dx}+17\frac{dy}{dx}-0=0\\ 6\times \left(2x\right)+3\left(1\times (y)+x\times \left(\frac{dy}{dx}\right)\right)+2\times \left(2y\frac{dy}{dx}\right)+17\frac{dy}{dx}=0\\ 12x+3y+\frac{dy}{dx}\left(3x+4y+17\right)=0\mathrm{.......}\left(2\right)\\ \text{Now}\text{put}\text{the}\text{given}\text{point}\left(x_1,y_1\right)=\left(-1,0\right)\text{in}\text{equation}\left(2\right)\\ 12\left(-1\right)+3\left(0\right)+\frac{dy}{dx}\left(3\left(-1\right)+4\left(0\right)+17\right)=0\\ -12+\frac{dy}{dx}\left(-3+17\right)=0\\ \frac{dy}{dx}\left(-14\right)=12\\ \frac{dy}{dx}=m=-\frac{6}{7}\end{array}$

Here $\frac{dy}{dx}=m=-\frac{6}{7}$ is the slope of the tangent.

The slope point form of the tangent line is

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}\left(x_1,y_1\right)=\left(-1,0\right)\text{and}m=-\frac{6}{7}\\ \left(y-0\right)=-\frac{6}{7}\left(x-\left(-1\right)\right)\\ 7\left(y\right)=-6\left(x+1\right)\\ 7y=-6x-6\\ 7y+6x+6=0\end{array}$

Hence the equation of tangent is $7y+6x+6=0$ .

b.

To determine

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

Answer to Problem 21E

The equation of normal to the curve is $6y-7x-7=0$ .

Explanation of Solution

Given information:The equation of the curve is $6x^2+3xy+2y^2+17y-6=0$ and the point is $\left(x_1,y_1\right)=\left(-1,0\right)$ .

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:

  $\begin{array}{l}\text{Since}m=-\frac{6}{7}\text{then}\text{the}\text{slope}\text{of}\text{normal}\text{is}{m}_1\text{=}-\frac{1}{m}=\frac{7}{6}\end{array}$

Then the equation of normal at $\left(x_1,y_1\right)=\left(-1,0\right)$ is

  $\begin{array}{l}\left(y-y_1\right)=m_1\left(x-x_1\right)\\ \left(y-0\right)=\frac{7}{6}\left(x-\left(-1\right)\right)\\ 6(y-0)=7\left(x+1\right)\\ 6y=7x+7\\ 6y-7x-7=0\end{array}$

Hence the equation of normal to the curve is $6y-7x-7=0$ .