JEE Main 2026 — 21 January, Evening Shift — Question 25

Mid point of PQ lies on $\mathrm{x}-\mathrm{y}+1=0$

$\frac{2 \cos \theta+\alpha}{2}-\frac{2 \sin \theta-5 \alpha-2}{2}+1=0$

$2 \cos \theta+\alpha-2 \sin \theta+5 \alpha+2+2=0$

$$\begin{gathered} \cos \theta-\sin \theta+3 \alpha+2=0 \end{gathered}$$

∵ Slope of PQ is $-1$

$\frac{2 \sin \theta+5 \alpha+2}{2 \cos \theta-\alpha}=-1$

$2 \sin \theta+5 \alpha+2=-2 \cos \theta+\alpha$

$\begin{gathered} \sin \theta+\cos \theta+2 \alpha+1=0 \end{gathered}$ eliminate $\alpha$ from and (2)

$\Rightarrow \cos \theta+5 \sin \theta=1, \theta \in[0,2 \pi]$

$\Rightarrow 5 \times 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}=2 \sin ^{2} \frac{\theta}{2}$

$\therefore \sin \frac{\theta}{2}=0 \Rightarrow \cos \theta=1$ or $\sin \frac{\theta}{2}=5 \Rightarrow \cos \theta=-\frac{12}{13}$

Sum of all possible values of abscissa of point P is $=2 \times 1+2\left(\frac{-12}{13}\right)=\frac{2}{13}$

$\therefore 13$ times sum of all possible values of abscissa of point P is 2 .