Straight Lines
Straight Line
star_batch_jee_advanced_2025
Grade 11
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 the line 'L':
If \frac{2}{OP} = \frac{1}{OA} + \frac{1}{OB}, then locus of P is 3y - 3x = 40
If OP^2 = (OA)(OB), then locus of P is (y - x)^2 = 200
If \frac{1}{OP^2} = \frac{1}{(OA)^2} + \frac{1}{(OB)^2}, then locus of P is (y - x)^2 = 80
If \frac{1}{OP^2} = \frac{1}{(OA)^2} + \frac{1}{(OB)^2}, then locus of P is (y - x)^2 = 80
Step-by-Step Solution
Key Concept: The locus of point P depends on the geometric relation used—whether it involves the arithmetic mean, geometric mean, or harmonic mean of the distances $OA$ and $OB$.
For option (A), using $\frac{2}{OP} = \frac{1}{OA} + \frac{1}{OB}$ with $OA = \frac{10}{\sin\theta - \cos\theta}$ and $OB = \frac{20}{\sin\theta - \cos\theta}$, we get $\frac{40}{r} = 3\sin\theta - 3\cos\theta$, which gives the locus $3y - 3x = 40$. For option (B), using $OP^2 = (OA)(OB)$ yields $r^2 = \frac{200}{(\sin\theta - \cos\theta)^2}$, so $(r\sin\theta - r\cos\theta)^2 = 200$, giving locus $(y-x)^2 = 200$. For option (C), using $\frac{1}{OP^2} = \frac{1}{OA^2} + \frac{1}{OB^2}$ produces the locus $(y-x)^2 = \frac{400}{3}$. For option (D), the harmonic mean relation yields locus $(y-x)^2 = 80$.
Correct Answer: 1,2,4