Q.A point on the hypotenuse of a triangle is at distance $a$ and $b$ from the sides of the triangle. Show that the minimum length of the hypotenuse is $(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$.

Idea: Let the hypotenuse make angle $\theta$ with one leg; write its length as a single-variable function of $\theta$ and minimise it.

The fixed point on the hypotenuse is at perpendicular distances $a$ and $b$ from the two legs. Splitting the hypotenuse at the foot of these perpendiculars gives its length as

Differentiate and set to zero:

So $\tan\theta=(a/b)^{1/3}$, giving

Substitute back: …

The set-up

A right triangle has a point $P$ on its hypotenuse that is at perpendicular distance $a$ from one leg and $b$ from the other. As the triangle changes shape (keeping $P$ at those fixed distances), the hypotenuse length changes; we want its smallest value.

Let $\theta$ be the angle the hypotenuse makes with the leg that is distance $a$ from $P$. Drop perpendiculars from $P$ to the two legs. These split the hypotenuse into two pieces:

• the piece near one end has length $\dfrac{a}{\sin\theta}$,
• the piece near the other end has length $\dfrac{b}{\cos\theta}$.

So the whole hypotenuse has length

This is already a single-variable function — exactly what we want to minimise.

Minimise with the derivative

Differentiate:

Set $L'(\theta)=0$:

Hence

Turn the angle into the length

From $(1)$, build a right triangle with "opposite" $=a^{1/3}$ and "adjacent" $=b^{1/3}$, so the hypotenuse of that reference triangle is $\sqrt{a^{2/3}+b^{2/3}}$. Then …

Method: Minimising a Length by Parametrising with an Angle

This method applies whenever a fixed point's distances from two perpendicular lines are given, and you need the minimum length of a segment (like a hypotenuse) through that point touching both lines — the natural variable is the angle the segment makes with one line, not a length.

Steps

Step 1: Identify the two fixed perpendicular distances and set up an angle $\theta$

Let $\theta$ be the angle the segment makes with one of the two lines. Dropping perpendiculars from the fixed point to each line splits the segment into two pieces, each expressible using $\theta$ and one of the given distances via right-triangle trigonometry.

Step 2: Write the total length as a single-variable trig function of $\theta$

Step 3: Differentiate and set $L'(\theta)=0$

Use $\dfrac{d}{d\theta}\csc\theta = -\csc\theta\cot\theta$ and $\dfrac{d}{d\theta}\sec\theta = \sec\theta\tan\theta$, then simplify the resulting equation into a single relation in $\tan\theta$.

Step 4: Solve for $\tan\theta$ and convert to $\sin\theta,\cos\theta$ …

Common Mistakes

Mistake 1: Treating the segment length as a function of a linear distance instead of the angle

Why it's wrong: trying to set up the problem with a linear variable (like the distance along one line) usually leads to an unnecessarily messy expression with square roots that resist clean differentiation; the angle parametrisation is what makes the trig derivatives collapse neatly. Correct approach: recognise "fixed point, two perpendicular distances, minimise the connecting segment" as the cue to parametrise by the angle the segment makes with one line.

Mistake 2: Losing track of which given distance pairs with $\sin\theta$ versus $\cos\theta$

Why it's wrong: swapping $a$ and $b$ between the sine and cosine terms swaps the two given distances' roles, producing intermediate expressions with $a$ and $b$ interchanged (harmless for this problem's symmetric final formula, but a real error in the intermediate steps and in any similar problem where the two distances play asymmetric roles). Correct approach: re-derive the right-triangle relation carefully from the picture, and stay consistent about which leg $\theta$ is measured from. …