Question

If $A(1,p^2);B(0,1)$ and $C(p,0)$ are the co ordinates of three points then the value of p for which the area of triangle ABC is minimum, is

• A. $\frac{1}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$ or $-\frac{1}{\sqrt{3}}$
• D. None

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

Given $A(1,p^2),B(0,1),C(p,0)$

Find perpendicular distance from A to BC, to find area $\left(△ABC\right)$

Equation of $BC=(y-0)=\frac{(0-1)}{(p-0)}(x-p)$

$py+x-p=0$

Perpendicular distance from A to BC$=\frac{|p\left(p^2\right)+\left(1\right)-p|}{\sqrt{p^2+1}}$

$=\frac{p^3-p+1}{\sqrt{p^2+1}}$

$BC=\sqrt{p^2+1}$

Area of $△ABC=\frac{1}{2}\times \left(\frac{p^3-p+1}{\sqrt{p^2+1}}\right)\left(\sqrt{p^2+1}\right)$

$A=\frac{p^3-p+1}{2}$

Area should be minimum, $\therefore \frac{d^{2}A}{d{p}^2}>0$ at p which satisfies $\frac{dA}{dp}=0$

$\frac{dA}{dp}=\frac{3p^2-1}{2}=0\Rightarrow p=\pm \frac{1}{\sqrt{3}}$

$\frac{d^{2}A}{d{p}^2}=\frac{6p}{2}=3p,A+p=\frac{1}{\sqrt{3}},\frac{d^{2}A}{d{p}^2}>0$

$\therefore$Area should be minimum for $p=\frac{1}{\sqrt{3}}$