Chapter 4.2, Problem 22E

a.

To determine

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

Answer to Problem 22E

The equation of tangent is $y=2$ .

Explanation of Solution

Given information:The equation of the curve is $x^2-\sqrt{3}xy+2y^2=5$ and the point is $\left(x_1,y_1\right)=\left(\sqrt{3},2\right)$ .

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 $x^2-\sqrt{3}xy+2y^2=5$ ............ (1)

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

By differentiating equation (1) we get

  $\begin{array}{l}{x}^2-\sqrt{3}xy+2y^2=5\\ \frac{d{x}^2}{dx}-\sqrt{3}\frac{d\left(xy\right)}{dx}+2\frac{d{y}^2}{dx}=0\\ 2x-\sqrt{3}\left(1\times (y)+x\times \left(\frac{dy}{dx}\right)\right)+2\times \left(2y\frac{dy}{dx}\right)=0\\ 2x-\sqrt{3}y+\frac{dy}{dx}\left(-\sqrt{3}x+4y\right)=0\mathrm{........}\left(2\right)\\ \text{Now}\text{put}\text{the}\text{given}\text{point}\left(x_1,y_1\right)=\left(\sqrt{3},2\right)\text{in}\text{equation}\left(2\right)\\ 2\left(\sqrt{3}\right)-\sqrt{3}\left(2\right)+\frac{dy}{dx}\left(-\sqrt{3}\left(\sqrt{3}\right)+4\left(2\right)\right)=0\\ 0+\frac{dy}{dx}\left(-3+8\right)=0\\ \frac{dy}{dx}=m=0\end{array}$

Here $\frac{dy}{dx}=m=0$ is the slope of the tangent.

The slope point form of the tangent line is

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}\left(x_1,y_1\right)=\left(\sqrt{3},2\right)\text{and}m=0\\ \left(y-2\right)=0\left(x-\left(\sqrt{3}\right)\right)\\ \left(y-2\right)=0\\ y=2\end{array}$

Hence the equation of tangent is $y=2$ .

b.

To determine

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

Answer to Problem 22E

The equation of normal to the curve is $y=2$ .

Explanation of Solution

Given information:The equation of the curve is $x^2-\sqrt{3}xy+2y^2=5$ and the point is $\left(x_1,y_1\right)=\left(\sqrt{3},2\right)$ .

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:

  $\begin{array}{l}\text{Since}m=0\text{then}\text{the}\text{slope}\text{of}\text{normal}\text{is}{m}_1\text{=}-\frac{1}{m}=0\end{array}$

Here $m_1=m=0$

Then the equation of normal at $\left(x_1,y_1\right)=\left(\sqrt{3},2\right)$ is same as the equation of tangent.

Hence the equation of normal to the curve is $y=2$ .