Coordinate Geometry Question 132
Question: Let P be the point (1, 0) and Q a point of the locus $ y^{2}=8x $ . The locus of midpoint of PQ is
[AIEEE 2005]
Options:
A) $ x^{2}+4y+2=0 $
B) $ x^{2}-4y+2=0 $
C) $ y^{2}-4x+2=0 $
D) $ y^{2}+4x+2=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ P=(1,0),Q=(h,k) $ such that $ k^{2}=8h $ Let $ (\alpha ,\beta ) $ be the midpoint of $ PQ $ ; $ \alpha =\frac{h+1}{2},\beta =\frac{k+0}{2};2\alpha -1=h,2\beta =k $
$ {{(2\beta )}^{2}}=8(2\alpha -1)\Rightarrow {{\beta }^{2}}=4\alpha -2\Rightarrow y^{2}-4x+2=0 $ .
BETA
