Straight Lines
Straight Line
star_batch_jee_advanced_2025
Grade None
Question:
A line passing through the origin $(O)$ has point $A$ and $B$ in the same direction such that $OA = AB = r$. Through points $A$ and $B$ two lines are drawn making equal angle $\tan^{-1}\left(\sqrt{3}\right)$ with the line $AB$. Then points which lies on the locus of point of intersection of the lines is/are:
$\left(r, \sqrt{2}r\right)$
$\left(\sqrt{2}r, r\right)$
$(r, r)$
$\left(-\sqrt{2}r, r\right)$
Step-by-Step Solution
Key Concept: The Pythagorean theorem on right triangle $OCP$ combined with the angle constraint $\alpha = \tan^{-1}\sqrt{3}$ determines the locus as a circle.
Given $\alpha = \tan^{-1}\sqrt{3}$, we have $\tan \alpha = \sqrt{3}$. From the geometry, $\tan \alpha = \frac{PC}{r/2}$, so $PC = \frac{r}{2}\tan \alpha = \frac{r\sqrt{3}}{2}$. Using the constraint $(OP)^2 = (OC)^2 + (CP)^2$ with $OA = AB = r$, we get $h^2 + k^2 = \left(\frac{3r}{2}\right)^2 + \left(\frac{r\sqrt{3}}{2}\tan\alpha\right)^2 = \left(\frac{9 + \tan^2\alpha}{4}\right)r^2$. The locus is $x^2 + y^2 = \left(\frac{9 + \tan^2\alpha}{4}\right)r^2$.
Correct Answer: 1,2,4