Chapter 4.4, Problem 65E

a.

To determine

Show that the area of the triangle is $A(x)=-f\text{'}\left(x\right)\left[x-\frac{f\left(x\right)}{f\text{'}\left(x\right)}\right]$ .

Answer to Problem 65E

Area of triangle $RST=-f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2$ .

Here,

  $f\left(x\right)=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}$and $f\text{'}(x)=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Explanation of Solution

Given:

Ellipse of equation: $\frac{x^2}{100}+\frac{{\left(y-5\right)}^2}{25}=1$

A triangle $RST$ is formed by using the tangent lines to the ellipse at $Q$ and $P$ .

  

Concept Used:

The equation of tangent line that passes through the point $P(x,a)$ is $y-y_0=m\left(x-x_0\right)$ .

Here, $m$ is the slope.

Calculation:

Rewrite the equation $\frac{x^2}{100}+\frac{{\left(y-5\right)}^2}{25}=1$ ,

  $\begin{array}{c}\frac{x^2}{100}+\frac{{\left(y-5\right)}^2}{25}=1\\ \frac{{\left(y-5\right)}^2}{25}=1-\frac{x^2}{100}\\ {\left(y-5\right)}^2=25\left(1-\frac{x^2}{100}\right)\\ y-5=5\sqrt{\left(1-\frac{x^2}{100}\right)}\\ y=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}\end{array}$

First Derivative:-

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(5+5\sqrt{\left(1-\frac{x^2}{100}\right)}\right)\\ y\text{'}=0+5\left(\frac{1}{2}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{-2x}{100}\right)\right)\\ y\text{'}=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Take $y=f(x)$ then $f\left(x\right)=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}$.

Also, $y\text{'}=f\text{'}(x)$ then $f\text{'}(x)=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Slope of ellipse :

  $\begin{array}{l}m=y\text{'}\\ m=f\text{'}(x)\end{array}$

Slope of ellipse at point $P(x,a)=\left(x_0,a\right)$ is $m=f\text{'}(x_0)$

Take point $P(x,a)=\left(x_0,a\right)$ ,

As, given that point $P$ lies on the ellipse then it’ll satisfy the equation of ellipse.

Put $x=x_0,y=a$

  $\begin{array}{c}y=f(x)\\ a=f(x_0)\end{array}$

Slope: - of ellipse passes through point $P$

  $\left(x=x_0,y=a\right)$

  $\begin{array}{c}y-y_0=m\left(x-x_0\right)\\ y-a=f\text{'}(x_0)\left(x-x_0\right)\end{array}$

The x -intercept of line: -

Since, the value of $y$ is zero on $x-$ axis.

For x −intercept put $y=0$ ,

  $\begin{array}{c}y-a=f\text{'}(x_0)\left(x-x_0\right)\\ 0-a=f\text{'}(x_0)\left(x-x_0\right)\\ \left(x-x_0\right)=\frac{-a}{f\text{'}(x_0)}\\ x=\left(x_0-\frac{a}{f\text{'}(x_0)}\right)\end{array}$

So, $x$ −intercept is $\left(x_0-\frac{a}{f\text{'}(x_0)},0\right)$ .

And the co-ordinates of vertex $T$ is $\left(x_0-\frac{a}{f\text{'}(x_0)},0\right)$ .

And, the distance of $T$ from origin $=OT$

And, $OT=x_0-\frac{a}{f\text{'}(x_0)}.$

The $y$ -intercept of line: -

Since, the value of $x$ is zero on $y-$ axis.

For $y$ −intercept put $x=0$ ,

  $\begin{array}{c}y-a=f\text{'}(x_0)\left(x-x_0\right)\\ y-a=f\text{'}(x_0)\left(0-x_0\right)\\ y=a-x_{0}f\text{'}(x_0)\\ y=-f\text{'}(x_0)\left(x_0-\frac{a}{f\text{'}(x_0)}\right)\\ y=-f\text{'}(x_0)\left(x_0-\frac{a}{f\text{'}(x_0)}\right)\end{array}$

So, $x$ −intercept is $\left(0,-f\text{'}(x_0)\left(x_0-\frac{a}{f\text{'}(x)}\right)\right)$ .

And the co-ordinates of vertex $R$ is $\left(0,-f\text{'}(x_0)\left(x_0-\frac{a}{f\text{'}(x_0)}\right)\right)$ .

The distance of $R$ from origin $=OR.$

Also, $OR=-f\text{'}(x_0)\left(x_0-\frac{a}{f\text{'}(x_0)}\right).$

Area:-

  

Area of triangle $OSR=$ Area of triangle $ORT$

So, Area of triangle $RST=$ Area of triangle $OSR+$ Area of triangle $ORT$

Area of triangle $RST=2\left(\text{Area of triangle }ORT\right)$ ,

Area of triangle $RST=A$ ,

  $\begin{array}{c}A=2\left(\frac{1}{2}\times \text{Base}\times \text{height}\right)\\ A=\left(x_0-\frac{a}{f\text{'}(x_0)}\right)\times \left(-f\text{'}(x_0)\left(x_0-\frac{a}{f\text{'}(x_0)}\right)\right)\\ A=-f\text{'}(x_0){\left(x_0-\frac{a}{f\text{'}(x_0)}\right)}^2\end{array}$

Area of triangle $RST=$

  $-f\text{'}(x_0){\left(x_0-\frac{a}{f\text{'}(x)}\right)}^2$ .

But, $f(x_0)=a$ then

Area of triangle $RST=-f\text{'}(x_0){\left(x_0-\frac{f\left(x_0\right)}{f\text{'}(x_0)}\right)}^2$ .

Put $x_0=x$ , in Area of triangle $RST$

Area of triangle $RST=$

  $-f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2$ .

Here,

  $f\left(x\right)=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}$

  $f\text{'}(x)=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Conclusion:

Area of triangle $RST=$

  $-f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2$ .

Here,

  $f\left(x\right)=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}$

  $f\text{'}(x)=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

b.

To determine

To Find: the domain of $A$ and draw the graph of $A$ . Find, how are the asymptotes of the graph related to the problem situation.

Explanation of Solution

Given:

Area of triangle $RST=-f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2$ .

Here,

  $f\left(x\right)=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}$

  $f\text{'}(x)=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Calculation:

Rewrite the area,

Area of triangle $RST=-f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2$ .

Here,

  $f\left(x\right)=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}$

  $f\text{'}(x)=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Area is defined if $A>0$then

  $\begin{array}{c}A>0\\ -f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2>0\\ \text{If -}f\text{'}(x)>0\\ -\left(-\frac{x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)>0\\ \frac{x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}>0\\ x>0\end{array}$

  $\begin{array}{c}\text{If }{\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2>0\\ {\left(\frac{xf\text{'}(x)-f\left(x\right)}{f\text{'}(x)}\right)}^2>0\\ xf\text{'}(x)>f\left(x\right)\\ \frac{x^2}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}>5+5\sqrt{\left(1-\frac{x^2}{100}\right)}\end{array}$

On solving,

  $x<10$ .

So, Domain: $\left(0,10\right)$ .

Graph:

  

Here, $X-axis$ denotes the values of $x$ and $Y-axis$ denotes the values of $A\left(x\right)$ .

According to graph, the asymptotes of graph does not form triangle as tangent lines formed in given problem.

The vertical asymptote is $x=0$ and Horizontal asymptote is $y=10$ .

c.

To determine

Find the height of the triangle with minimum area.

Also, check if it is related to the y-coordinate of the center of the ellipse.

Answer to Problem 65E

Height $=15$ units.

Also, height is related to the y-coordinate of the center of the ellipse. It is the triple time of the y-coordinate of the center of the ellipse.

Explanation of Solution

Given:

Area of triangle $A(x)=-f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2$ .

Here,

  $f\left(x\right)=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}$

  $f\text{'}(x)=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Calculation:

Differentiate the $A(x)$ with respect to $x$ ,

  $A\text{'}(x)=-f\text{'}\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2-2f\text{'}(x)\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)\left(1-\frac{{\left(f\text{'}(x)\right)}^2-f\left(x\right)f\text{'}\text{'}(x)}{f\text{'}(x)}\right)$

For point of minima put $A\text{'}(x)=0$ and solve.

  $x=5\sqrt{3},a=2.5$ .

Height $OT=x_0-\frac{a}{f\text{'}(x_0)}.$

Put the value and solve,

Height $=15$ units.

Conclusion:

Height $=15$ units.

Also, height is related to the y-coordinate of the center of the ellipse. It is the triple time of the y-coordinate of the center of the ellipse.

d.

To determine

Repeat the part (a) to (c) for the ellipse $\frac{x^2}{C^2}+\frac{{(y-B)}^2}{B^2}=1$ .

Answer to Problem 65E

Area of triangle $A(x)=-f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2$ .

Domain: $\left(0,C\right)$ .

Explanation of Solution

Given:

From part (a)

Area of triangle $A(x)=-f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2$ .

Here,

  $f\left(x\right)=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}$

  $f\text{'}(x)=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Calculation:

Solve in the same way as part (a), (b) and (c).

Area of triangle $A(x)=-f\text{'}(x){\left(x-\frac{f\left(x\right)}{f\text{'}(x)}\right)}^2$ .

Here,

  $f\left(x\right)=5+5\sqrt{\left(1-\frac{x^2}{100}\right)}$

  $f\text{'}(x)=\frac{-x}{20}{\left(1-\frac{x^2}{100}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Domain: $(0,C)$

Since, Graph Is same for both parts. So,

Graph:

  

Here, $X-axis$ denotes the values of $x$ and $Y-axis$ denotes the values of $A\left(x\right)$ .