Question

A variable line $L$ is drawn through $O(0,0)$ to meet the lines$L_1:Y-X-10=0$ And $L_2:Y-X-20=0$ . At points $A$ And $B$ Respectively. A Point $P$ is taken on $L$ such that $\frac{2}{OP}=\frac{1}{OA}+\frac{1}{OB}$ And $P,A,B$ lies on the same side of origin$O$. The locus of P is

Answer 1

Let, the given value of $x\&y$ are,

$$\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}$$

Now, by substituting in $L_1$ we get,

$\frac{1}{OA}=\frac{\sin \theta -\cos \theta }{10}$

Now, by substituting in$L_2$, we get,

$\frac{1}{OB}=\frac{sin\theta -\cos \theta }{20}$

Therefore,

$\frac{2}{r}=\frac{\sin \theta -\cos \theta }{10}+\frac{\sin \theta -\cos \theta }{20}$

$$\begin{gathered} 40=3\left(r\sin \theta \right)-3\left(\cos \theta \right) \\ =3y-3x \end{gathered}$$

Therefore, the Locus of $P$ is $3y-3x=40$