Conic-Sections Question 558
Question: If tangents are drawn to the parabola $ y^{2}=4ax $ at points whose abscissae are in the ratio $ m^{2}:1, $ then the locus of their point of intersection is the curve $ ( m>0 ) $
Options:
A) $ y^{2}={{({m^{1/2}}-{m^{-1/2}})}^{2}}ax $
B) $ y^{2}={{({m^{1/2}}+{m^{-1/2}})}^{2}}ax $
C) $ y^{2}={{({m^{1/2}}+{m^{-1/2}})}^{2}}x $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Consider two points $ P( at^2_1,2at_1 ) $ and $ Q( at^2_2,2at_2 ) $ on the parabola $ y^{2}=4ax $ Given: $ \frac{at^2_1}{at^2_2}=\frac{m^{2}}{1} $ or $ t_1=mt_2 $ ? (1) Let $ R(h,k) $ be the point of intersection of tangents at P and Q. Then, $ h=at_1t_2 $ and $ k=a(t_1+t_2) $
$ \Rightarrow h=amt^2_2 $ and $ k=a(mt_2+t_2) $ [Using (1)]
$ \Rightarrow t^2_2=\frac{h}{am} $ and $ t_2=\frac{k}{a(m+1)} $ Equating the two values of $ t_2, $ we get $ \frac{k^{2}}{a^{2}{{(m+1)}^{2}}}=\frac{h}{am} $
$ \Rightarrow k^{2}=ah\frac{{{(m+1)}^{2}}}{m}\Rightarrow k^{2}=ah{{( \sqrt{m}+\frac{1}{\sqrt{m}} )}^{2}} $
$ \therefore $ Required locus is $ y^{2}=ax({m^{\frac{1}{2}}}+{m^{\frac{1}{2}}}). $
BETA
