Conic Sections
Conic Section
star_batch_jee_advanced_2025
Grade 11
Question:
The straight line $ax + by + c = 0$ cuts the locus of point of intersection of the lines $\frac{tx}{3} + t = 0, \frac{x}{4} + \frac{ty}{3} - t = 0$ at $A$ & $B$ such that line $AB$ subtends a right angle at the origin, then $\left[\frac{3a - 4b}{c}\right]$ is _____. ($[|$ represents greatest integer function$)$
Step-by-Step Solution
Key Concept: Eliminating the parameter from two variable lines gives the locus, and perpendicularity of resulting lines imposes a constraint on coefficients $a, b, c$.
Given two lines $\frac{x}{4} - \frac{y}{3} + t = 0$ and $\frac{x}{4} + \frac{y}{3} - t = 0$, their locus of intersection is found by eliminating $t$: $t = \frac{3-x}{\frac{x}{4}+1} = \frac{x-4}{1-\frac{y}{3}}$. The curve $\frac{y}{3} - \frac{y^2}{9} - \frac{x^2}{16} + \frac{x}{4} = 0$ represents equation (3). The pair of straight lines from the origin to intersection points of $ax + by + c = 0$ and (3) are given by $\left(\frac{x^2}{16} - \frac{y^2}{9}\right)\left(\frac{x}{4} - \frac{y}{3}\right)\left(\frac{ax+by}{-c}\right) = 0$. For perpendicular lines, $\frac{c-4a}{16c} + \frac{c+3b}{9c} = 0$, leading to $36a - 48b - 25c = 0$ and thus $\left[\frac{3a-4b}{c}\right] = \left[\frac{25}{12}\right] = 2$.
Correct Answer: 2