MCQ
If two perpendicular rays from focus of parabolic surface $y^2 = 4x$ are incident at points $A(t^2_1,2t_1)$ & $A(t^2_2,2t_2)$ such that $t_1t_2 = -10$, then distance between the reflected rays will be -
- A$9$
- B$6$
- ✓$18$
- Dnot a constant
✓
Answer
Correct option: C.
$18$
c
Reflected rays will be two parallel lines and distance between them $=2\left|t_{1}-t_{2}\right|$
Reflected rays will be two parallel lines and distance between them $=2\left|t_{1}-t_{2}\right|$
also $As$ $\perp \mathrm{BS} \Rightarrow\left(\frac{2 \mathrm{t}_{1}}{\mathrm{t}_{1}^{2}-1}\right)\left(\frac{2 \mathrm{t}_{2}}{\mathrm{t}_{2}^{2}-1}\right)=-1$
$\Rightarrow\left(t_{1}-t_{2}\right)^{2}=\left(1+t_{1} t_{2}\right)^{2} $
$\Rightarrow\left|t_{1}-t_{2}\right|=9$
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