Conic Sections Question 68
Question: If the chord joining the points $ (at_1^{2},\ 2at_1) $ and $ (at_2^{2},\ 2at_2) $ of the parabola $ y^{2}=4ax $ passes through the focus of the parabola, then
[MP PET 1993]
Options:
A) $ t_1t_2=-1 $
B) $ t_1t_2=1 $
C) $ t_1+t_2=-1 $
D) $ t_1-t_2=1 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{(y-2at_2)}{(2at_2-2at_1)}=\frac{x-at_2^{2}}{(at_2^{2}-at_1^{2})} $ ; As focus i.e., (a, 0) lies on it,
therefore $ \frac{-2at_2}{2a(t_2-t_1)}=\frac{a(1-t_2^{2})}{a(t_2-t_1)(t_2+t_1)} $
therefore $ -t_2=\frac{(1-t_2^{2})}{(t_2+t_1)} $
therefore $ -t_2^{2}-t_1t_2=1-t_2^{2} $
therefore $ t_1t_2=-1 $ .
BETA
