I am given the following question, I have had a several gos on it but didnt manage to solve it at all.
Question givne the function $$f(x)=\frac{1}x, x>0$$ given an arbitrary point on the graph bounded by the tangent and the normal lines to the point which together with the x-axis they form a right angle triangle, what is the smallest length that can the hypotenuse obtain?
Hint the answer is : $\frac {4}{3^\frac{3}4}$
Best Answer
Here's a long solution.....
Derivative of function is $\frac{-1}{x^2}$
Let us consider a point $x=p$ so $f(x)=1/p$
Now the equation of tangent line at $x=a$
$$y-\frac1p=\frac{-1}{p^2}(x-p)$$
Equation of normal line is $$y-\frac1p=p^2(x-p)$$
The $x$-intercepts of both lines are $$x=2p$$ and $$x=p-\frac{1}{p^3}$$ respectively.
Finding their distance to $p$ using distance formula gives two sides lengths.
Side $1$=$$\sqrt{p^2+\frac{1}{p^2}}$$ and Side $2$=$$\sqrt{\frac{1}{p^6}+\frac{1}{p^2}}$$ Since (By Pythagoras Theorem) $$(Side \ 1)^2+(Side \ 2)^2=(Hypotenuse)^2$$
SO, hypotenuse $=$ $$\sqrt{\frac{1}{p^6}+\frac{2}{p^2}+p^2}$$ Computing derivative $$\frac{\frac{-6}{p^7}-\frac{4}{p^3}+2p}{2\sqrt{\frac{1}{p^6}+\frac{2}{p^2}+p^2}}$$ and setting it to zero gives $$\frac{-6}{p^7}-\frac{4}{p^3}+2p=0$$ which equals $$p^8-2p^4-3=0$$ which when factorised gives
$$(p^4-3)(p^4+1)=0$$ which gives $p=\sqrt[4]3$ Putting this in $$\sqrt{\frac{1}{p^6}+\frac{2}{p^2}+p^2}$$ and simplifying gives...
$$Hypotenuse=\frac{4}{3^{\frac34}}$$
EDIT:
As noted in comments the hypotenuse is simply $$2p-p+\frac{1}{p^3}=p+\frac{1}{p^3}$$ After taking derivative and setting it to zero easily gives $$p=\sqrt[4]3$$