Conic Sections Question 155
Question: The length of the normal chord to the parabola $ y^{2}=4x $ , which subtends right angle at the vertex is
[RPET 1999]
Options:
A) $ 6\sqrt{3} $
B) $ 3\sqrt{3} $
C) 2
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
Normal at $ P(t_1^{2},2t_1) $ on the parabola $ y^{2}=4x $ …..(i) Meets it again at the point $ (y-12)=\frac{-36}{54}(x+36)\Rightarrow 2x+3y+36=0 $ , where $ t_2=-t_1-\frac{2}{t_1} $ ……(ii) If $ PQ $ subtends a right angle at the vertex (0, 0) then (Slope of OP) (Slope of $ OQ) $
$ =-1 $
$ \Rightarrow $
$ \frac{2t_1}{t_1^{2}}.\frac{2t_2}{t_2^{2}}=-1 $
$ \Rightarrow $ $ t_2=-\frac{4}{t_1} $ ……(iii) From (ii) and (iii), $ -t_1-\frac{2}{t_1}=-\frac{4}{t_1} $
$ \Rightarrow $ $ -t_1=-\frac{2}{t_1} $
$ \Rightarrow t_1^{2} $ = 2
$ \Rightarrow t_1=\pm \sqrt{2} $ ;
$ \therefore t_2=\mp 2\sqrt{2} $
$ \therefore $ $ P $ and $ Q $ are $ (2,\pm 2\sqrt{2}) $ and $ (8,\mp 4\sqrt{2}) $
$ \therefore $ $ PQ=\sqrt{{{(8-2)}^{2}}+{{(\mp 4\sqrt{2}\mp 2\sqrt{2})}^{2}}}=\sqrt{36+72} $
$ =\sqrt{108}=6\sqrt{3.} $
BETA
