IIT-JEE 1995Question 1
Let $(h,k)$ be a fixed point, where $h>0,k>0$. A straight line passing through this point cuts the positive direction of the coordinate axes at the points $P$ and $Q$. Find the minimum area of the triangle $OPQ$, $O$ being the origin.
Differentiation and its Application
Answer$2hk$
Solution
Concept. Express the area by one intercept and minimise. Formulas used. (1) Line $\frac xa+\frac yb=1$ through $(h,k)$: $\frac ha+\frac kb=1$Step 1 - Area in $a$. By (1), $b=\frac{ka}{a-h}$ ($a>h$), so $$A(a)=\frac12ab=\frac{ka^2}{2(a-h)}.$$ Step 2 - Differentiate. $$A'(a)=\frac k2\cdot\frac{2a(a-h)-a^2}{(a-h)^2}=\frac{ka(a-2h)}{2(a-h)^2}.$$ $A'<0$ for $a<2h$ and $A'>0$ for $a>2h$: minimum at $a=2h$. Step 3 - Value. $b=2k$, area $=\frac12\cdot2h\cdot2k=2hk$. Answer: $2hk$