Conic Sections Question 123
Question: The locus of the middle points of the chords of the parabola $ y^{2}=4ax $ which passes through the origin
[RPET 1997; UPSEAT 1999]
Options:
A) $ y^{2}=ax $
B) $ y^{2}=2ax $
C) $ y^{2}=4ax $
D) $ x^{2}=4ay $
Show Answer
Answer:
Correct Answer: B
Solution:
Any line through origin (0,0) is $ y=mx $ . It intersects $ y^{2}=4ax $ in $ ( \frac{4a}{m^{2}},\frac{4a}{m} ) $ . Mid point of the chord is $ ( \frac{2a}{m^{2}},\frac{2a}{m} ) $
$ x=\frac{2a}{m^{2}}, $
$ y=\frac{2a}{m} $
therefore $ \frac{2a}{x}=\frac{4a^{2}}{y^{2}} $ or $ y^{2}=2ax $ , which is a parabola.
BETA
