Conic Sections Question 134
Question: The locus of the midpoint of the line segment joining the focus to a moving point on the parabola $ y^{2}=4ax $ is another parabola with the directrix
[IIT Screening 2002]
Options:
A) $ x=-a $
B) $ x=-\frac{a}{2} $
C) $ x=0 $
D) $ x=\frac{a}{2} $
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Answer:
Correct Answer: C
Solution:
$ \alpha =\frac{at^{2}+a}{2},\beta =\frac{2at+0}{2}\Rightarrow 2\alpha =at^{2}+a,at=\beta $
$ 2\alpha =a.\frac{{{\beta }^{2}}}{a^{2}}+a $ or $ 2a\alpha ={{\beta }^{2}}+a^{2} $
The locus is $ y^{2}=\frac{4a}{2}( x-\frac{a}{2} ) $
$ =4b(x-b),( b=\frac{a}{2} ) $
Directrix is $ (x-b)+b=0 $ or $ x=0 $ .
BETA
