Chapter 7, Problem 12P

To determine

To find : the curve that passes through the point $\left(3,2\right)$

Answer to Problem 12P

The curve that passes through the point $\left(3,2\right)$ is $y=\frac{6}{x}$

Explanation of Solution

Given information : The question is

If the tangent line is drawn at any point $P$ on the curve, then the part of the tangent line that lies in the first quadrant is bisected at $P$

Calculation :

Assume that the curve follows $y=f\left(x\right)$ and a tangent is drawn at $x=a$

Slope of the tangent is $f\text{'}\left(a\right)$ and it passes through $\left(a,f\left(a\right)\right)$

Equation of tangent is

  $\begin{array}{l}y-f\left(a\right)=f\text{'}\left(a\right)\cdot \left(x-a\right)\\ y=f\text{'}\left(a\right)\cdot x-a\cdot f\text{'}\left(a\right)+f\left(a\right)\\ -f\text{'}\left(a\right)\cdot x+y=-a\cdot f\text{'}\left(a\right)f\left(a\right)\\ \frac{-f\text{'}\left(a\right)\cdot x}{-a\cdot f\text{'}\left(a\right)f\left(a\right)}+\frac{y}{-a\cdot f\text{'}\left(a\right)f\left(a\right)}=1\\ \frac{x}{a-\frac{f\left(a\right)}{f\text{'}\left(a\right)}}+\frac{y}{-a\cdot f\text{'}\left(a\right)f\left(a\right)}=1\end{array}$

The $x\&y$ intercepts are $a-\frac{f\left(a\right)}{f\text{'}\left(a\right)}$ and $-a\cdot f\text{'}\left(a\right)f\left(a\right)$ respectively

Since $\left(a,f\left(a\right)\right)$ is the midpoint of two intercepts

  $\begin{array}{l}a-\frac{f\left(a\right)}{f\text{'}\left(a\right)}=2a\Rightarrow \frac{f\left(a\right)}{f\text{'}\left(a\right)}=a\Rightarrow f\text{'}\left(a\right)=-\frac{f\left(a\right)}{a}\\ -a\cdot f\text{'}\left(a\right)f\left(a\right)=2f\left(a\right)\Rightarrow -a\cdot f\text{'}\left(a\right)=f\left(a\right)\Rightarrow f\text{'}\left(a\right)=-\frac{f\left(a\right)}{a}\\ f\text{'}\left(x\right)=-\frac{f\left(x\right)}{x}\\ \frac{dy}{dx}=-\frac{y}{x}\\ \frac{dy}{y}=-\frac{dx}{x}\end{array}$

Integrate both sides

  $\begin{array}{l}{\displaystyle \int \frac{dy}{y}=-{\displaystyle \int \frac{dx}{x}}}\\ \ln \left|y\right|=-\ln \left|x\right|+C\\ \ln \left|y\right|+\ln \left|x\right|=C\\ \ln \left|xy\right|=C\\ \ln \left|3\cdot 2\right|=C\\ \ln 6=C\\ \ln \left|xy\right|=\ln 6\\ e^{\ln \left|xy\right|}=e^{\ln 6}\\ \left|xy\right|=6\\ y=\pm \frac{6}{x}\end{array}$

Since $y=-\frac{6}{x}$ is never in first quadrant so rejected this case

So, $y=\frac{6}{x}$