Coordinate Geometry
Family of concurrent lines; max distance between two points
MMTS_Full_Test_15
Grade 12
Question:
Let $P$ be any point on $x-y+3=0$ and $A=(3,4)$. If the family of lines $(3\sec\theta+5\csc\theta)x+(7\sec\theta-3\csc\theta)y+11(\sec\theta-\csc\theta)=0$ are concurrent at $B$ for all permissible $\theta$, then max value of $|PA-PB|$ is
(A) 3
(B) $2\sqrt{10}$
(C) $2\sqrt{34}$
(D) 5
Step-by-Step Solution
Key Concept: Find $B$ by solving two linearly independent equations from the family. $|PA-PB|\leq|AB|$ with max at collinear config.
Max $|PA-PB|=2\sqrt{10}$.
Correct Answer: (B) $2\sqrt{10}$