Then locus of $P$ is?A point $P$ is taken on $'L'$ such that $\frac{2}{OP}=\frac{1}{OA}+\frac{1}{OB}$. Then locus of $P$ is?

Question

A variable straight line $'L'$ drawn through $O(0,0)$ to meet the lines $L_1=y-x-10=0$ and $y-x-20=0$ at point $A$ and $B$ respectively. A point $P$ is taken on $'L'$ such that $\frac{2}{OP}=\frac{1}{OA}+\frac{1}{OB}$. Then locus of $P$ is?

(A) $3x+3y-40=0$ (B) $3x+3y+40=0$ (C) $3x-3y-40=0$ (D) $3x-3y+40=0$

My Answer

I am getting locus of $P$ as $3x-3y+40$ and $3x-3y-40=0$ both as an answer but answer given is only $3x-3y+40=0$.

Is there anything I am misunderstanding? or answer given is wrong?

Answer 1

I think you are correct.

Let $P(X,Y)$.

• If $L:x=0$, then $A(0,10),B(0,20)$, so $\frac{2}{\sqrt{Y^2}}=\frac{1}{10}+\frac{1}{20}$ implies $P(0,\pm\frac{40}{3})$.

• If $L:y=ax\ (a\not=1)$, then $A(\frac{10}{a-1},\frac{10a}{a-1}),B(\frac{20}{a-1},\frac{20a}{a-1})$, so $\frac{2}{|X|\sqrt{1+a^2}}=\frac{|a-1|}{10\sqrt{1+a^2}}+\frac{|a-1|}{20\sqrt{1+a^2}}$ implies $P(\pm\frac{40}{3|a-1|},\pm\frac{40a}{3|a-1|})$

Therefore, we get $$X-Y=\pm\frac{40}{3|a-1|}-\bigg(\pm\frac{40a}{3|a-1|}\bigg)=\mp\frac{40}{3}\cdot\frac{a-1}{|a-1|}=\mp\frac{40}{3}$$