JEE Advanced 2014 Paper 2Question 1
The value of $r$ is
Parabola
Solution
Concept. (Passage: $P(at^2,2at)$, $Q$, $R(ar^2,2ar)$, $S(as^2,2as)$ lie on $y^2=4ax$; $PQ$ is a focal chord and $QR\parallel PK$ with $K(2a,0)$.) Formulas used. (1) Ends of a focal chord: $t_1t_2=-1$(2) Slope of the chord joining $t_1,t_2$ on $y^2=4ax$: $\frac{2}{t_1+t_2}$Step 1 - Point $Q$, use (1). $Q$ has parameter $-\frac1t$: $Q=\big(\frac a{t^2},-\frac{2a}t\big)$. Step 2 - Slope of $PK$. $$m_{PK}=\frac{2at-0}{at^2-2a}=\frac{2t}{t^2-2}$$ Step 3 - Slope of $QR$, use (2). Parameters $-\frac1t$ and $r$: $$m_{QR}=\frac{2}{r-\frac1t}$$ Step 4 - Parallel. $$\frac{2}{r-\frac1t}=\frac{2t}{t^2-2}\Rightarrow r-\frac1t=\frac{t^2-2}{t}\Rightarrow r=\frac{t^2-1}{t}$$ Answer: D