Chapter 4.2, Problem 19E

a.

To determine

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

Answer to Problem 19E

The equation of tangent is $y-3x-6=0$ .

Explanation of Solution

Given information:The equation of the curve is $x^2\times y^2=9$ and the point is $\left(x_1,y_1\right)=\left(-1,3\right)$ .

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 $x^2\times y^2=9$ ................(1)

And the point at which the tangent equation is to be determined is $\left(x_1,y_1\right)=\left(-1,3\right)$ .

By differentiating equation (1) we get

  $\begin{array}{l}{x}^2\times y^2=9\\ \frac{d\left(x^2\times y^2\right)}{dx}=0\\ 2x\times \left(y^2\right)+x^2\times \left(2y\frac{dy}{dx}\right)=0\mathrm{.......}\left(2\right)\\ \text{Now}\text{put}\text{the}\text{given}\text{point}\left(x_1,y_1\right)=\left(-1,3\right)\text{in}\text{equation}\left(2\right)\\ 2\left(-1\right)\times {\left(3\right)}^2+{\left(-1\right)}^2\times 2\left(3\right)\frac{dy}{dx}=0\\ -18+6\frac{dy}{dx}=0\\ \frac{dy}{dx}=m=3\end{array}$

Here $\frac{dy}{dx}=m=3$ is the slope of the tangent.

The slope point form of the tangent line is

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}\left(x_1,y_1\right)=\left(-1,3\right)\text{and}m=3\\ \left(y-3\right)=3\left(x-\left(-1\right)\right)\\ \left(y-3\right)=3\left(x+1\right)\\ y-3=3x+3\\ y-3x-6=0\end{array}$

Hence the equation of tangent is $y-3x-6=0$ .

b.

To determine

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

Answer to Problem 19E

The equation of normal to the curve is $3y+x-8=0$ .

Explanation of Solution

Given information:The equation of the curve is $x^2\times y^2=9$ and the point is $\left(x_1,y_1\right)=\left(-1,3\right)$ .

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:

  $\begin{array}{l}\text{Since}m=3\text{then}\text{the}\text{slope}\text{of}\text{normal}\text{is}{m}_1\text{=}-\frac{1}{m}=-\frac{1}{3}\end{array}$

Then the equation of normal at $\left(x_1,y_1\right)=\left(-1,3\right)$ is

  $\begin{array}{l}\left(y-y_1\right)=m_1\left(x-x_1\right)\\ \left(y-3\right)=-\frac{1}{3}\left(x-\left(-1\right)\right)\\ 3\left(y-3\right)=-1\left(x+1\right)\\ 3y-9=-x-1\\ 3y+x-8=0\end{array}$

Hence the equation of normal to the curve is $3y+x-8=0$ .