ABC is a right angled triangled, right angled at C and P is the length of the perpendicular from C on AB. If a, b and c are the length of the sides BC, CA and AB respectively, then

Answer: Option B

According to question,
ACB is a right angle triangle
∴ area of ΔACB

$$\frac{1}{2}$$ × AC × BC = $$\frac{1}{2}$$ × AB × PC
$$\frac{1}{2}$$ × b × a = $$\frac{1}{2}$$ × c × p
c = $$\frac{ab}{p}$$ . . . . . . . (i)
By using pythagoras theorem
AB² = AC² + BC²
c² = b² + a² . . . . . . . . . . (ii)
Put the value of C in equation (ii)
$$\begin{array}{rl}& {\left(\frac{ab}{p}\right)}^2=a^2+b^2\\ & \Rightarrow \frac{a^2{b}^2}{p^2}=a^2+b^2\\ & \Rightarrow \frac{1}{p^2}=\frac{a^2}{a^2{b}^2}+\frac{b^2}{a^2{b}^2}\\ & \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\end{array}$$

---

Alternate :
From figure,
$$\begin{array}{rl}& P=\frac{ab}{c}\\ & P=\frac{ab}{\sqrt{a^2+b^2}}\left(\because a^2+b^2=c^2\right)\\ & P^2=\frac{a^2{b}^2}{a^2+b^2}\\ & \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\end{array}$$