Question

Find the equation of the normal of the curve at the given points :D thankiewss

Answer 1

We want to find the equation of the normal line of $y=\dfrac{20-x}{3x}$ at the point $P=(x_0,y_0)$, where $y_0=3$. First calculate $x_0$. We have:

$y_0=f(x_0)=\dfrac{20-x_0}{3x_0}\\\\\\3=\dfrac{20-x_0}{3x_0}\quad|\cdot3x_0\\\\\\3\cdot3x_0=20-x_0\\\\9x_0=20-x_0\\\\9x_0+x_0=20\\\\10x_0=20\quad|:2\\\\\boxed{x_0=2}$

Now, when we know that $P=(x_0,y_0)=(2,3)$ we can write an equation of the normal line as:

$\boxed{y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)}$

Calculate $f'(x_0)$:

$f'(x)=\left(\dfrac{20-x}{3x}\right)'=\left(\dfrac{20}{3x}\right)'-\left(\dfrac{x}{3x}\right)'=\dfrac{20}{3}\cdot\left(\dfrac{1}{x}\right)'-\left(\dfrac{1}{3}\right)'=\\\\\\=\dfrac{20}{3}\cdot\left(-\dfrac{1}{x^2}\right)-0=\boxed{-\dfrac{20}{3x^2}}\\\\\\\\f'(x_0)=f'(2)=-\dfrac{20}{3\cdot2^2}=-\dfrac{20}{3\cdot4}=-\dfrac{20}{12}=\boxed{-\frac{5}{3}}$

and the equation of the normal line:

$y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\y-3=-\dfrac{1}{-\frac{5}{3}}(x-2)\\\\\\y-3=\dfrac{3}{5}(x-2)\\\\\\y-3=\dfrac{3}{5}x-\dfrac{6}{5}\\\\\\y=\dfrac{3}{5}x-\dfrac{6}{5}+3\\\\\\\boxed{y=\dfrac{3}{5}x+\dfrac{9}{5}}$