Question

Let $P\equiv (a,b),Q\equiv (c,d)$ and 0 < a < b < c < d $L\equiv (a,0),M\equiv (c,0)$, R lies on x axis such that PR + RQ is minimum then R divides LM

• A. internally in the ratio a : b
• B. internally in the ratio b : c
• C. internally in the ratio b : d
• D. internally in the ratio c : d

Answer 1

The correct option is C internally in the ratio b : d

$p\equiv (a,b),Q\equiv (c,d);0<a<b<c<dL\equiv (a,0),M\equiv (c,0)$

The point R is on x-axis such that PR+RQ is minimum

For PR+RQ to be minimum, $P,Q,R$ would have to b a straight line.But R lies on the x-axis.

By Fermat's principle of minimum time of path $PR+RQ$ as a function of point R for independent variable it can be shown by differentiation that the line joining image P and Q intersects x-axis at the required bounce off point R.

$\Rightarrow \frac{y-b}{d-b}=\frac{x-a}{c-a}\Rightarrow (y-b)(c-a)=(x-a)(d-b)\therefore y(c-a)-x(d-b)=bc-ad\therefore PR+RQ\text{is minimum}$

As ,$L\equiv (a,0),M\equiv (c,o)$

R divides LM in the ratio $b:d$