Chapter 4.2, Problem 17E

a.

To determine

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

Answer to Problem 17E

The equation of tangent is $4y-7x+2=0$ .

Explanation of Solution

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

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 $x^2+xy-y^2=1$ ................(1)

And the point at which the tangent equation is to be determined is $\left(x_1,y_1\right)=\left(2,3\right)$

By differentiating equation (1),

  $\begin{array}{l}{x}^2+xy-y^2=1\\ \frac{d{x}^2}{dx}+\frac{d\left(xy\right)}{dx}-\frac{d{y}^2}{dx}=\frac{d1}{dx}\\ 2x+\left(y+x\times \frac{dy}{dx}\right)-2y\frac{dy}{dx}=0\\ 2x+y+\frac{dy}{dx}\left(x-2y\right)=0\mathrm{......}(2)\\ \text{Now}\text{put}\text{the}\text{given}\text{point}\left(x_1,y_1\right)=\left(2,3\right)\text{in}\text{equation}\text{(2),}\\ 2\left(2\right)+3+\frac{dy}{dx}\left(2-2\left(3\right)\right)=0\\ \frac{dy}{dx}\left(-4\right)=-7\\ \frac{dy}{dx}=m=\frac{7}{4}\end{array}$

Here $\frac{dy}{dx}=m=\frac{7}{4}$ is the slope of the tangent.

The slope point form of the tangent line is:

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}\left(x_1,y_1\right)=\left(2,3\right)\text{and}m=\frac{7}{4}\\ \left(y-3\right)=\frac{7}{4}\left(x-2\right)\\ 4\left(y-3\right)=7\left(x-2\right)\\ 4y-12=7x-14\\ 4y-7x+2=0\end{array}$

Hence the equation of tangent is $4y-7x+2=0$ .

b.

To determine

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

Answer to Problem 17E

The equation of normal to the curve is $7y+4x-29=0$ .

Explanation of Solution

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

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:

  $\begin{array}{l}\text{Since}m=\frac{7}{4}\text{then}\text{the}\text{slope}\text{of}\text{normal}\text{is}{m}_1\text{=}-\frac{1}{m}=-\frac{4}{7}\end{array}$

Then the equation of normal at $\left(x_1,y_1\right)=\left(2,3\right)$ is

  $\begin{array}{l}\left(y-y_1\right)=m_1\left(x-x_1\right)\\ \left(y-3\right)=-\frac{4}{7}\left(x-2\right)\\ 7\left(y-3\right)=-4\left(x-2\right)\\ 7y-21=-4x+8\\ 7y+4x-29=0\end{array}$

Hence the equation of normal to the curve is $7y+4x-29=0$ .