Conic Sections Question 2
Question: The normal at the point $ (bt_1^{2},2bt_1) $ on a parabola meets the parabola again at the point $ (bt_2^{2},2bt_2) $ then
Options:
A) $ t_2=-t_1-\frac{2}{t_1} $
B) $ t_2=-t_1+\frac{2}{t_1} $
C) $ t_2=t_1-\frac{2}{t_1} $
D) $ t_2=t_1+\frac{2}{t_1} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given point on parabola is $ P(bt_1^{2},2bt_1) $ So consider parabola $ y^{2}=4bx $ Differentiating W.r.t.x, we get $ 2y\frac{dy}{dx}=4b $
$ \therefore \frac{dy}{dx}=\frac{2b}{y} $
$ \therefore $ slope of normal at point $ p=-\frac{2bt_1}{2b}=-t_1 $ . Also normal meets the curve again at point $ Q(bt^2_2,2bt_2) $ So slope of line $ PQ=\frac{2bt_2-2btl1}{bt^2_2-bt^2_1}=\frac{2}{t_1+t_2} $
$ \therefore -t_1=\frac{2}{t_1+t_2} $
$ \therefore t_2=-t_1-\frac{2}{t_1} $
BETA
