Conic Sections Question 15
Question: The equation of the locus of a point which moves so as to be at equal distances from the point (a, 0) and the y-axis is
Options:
A) $ y^{2}-2ax+a^{2}=0 $
B) $ y^{2}+2ax+a^{2}=0 $
C) $ x^{2}-2ay+a^{2}=0 $
D) $ x^{2}+2ay+a^{2}=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Accordingly, $ {{(h-a)}^{2}}+k^{2}=h^{2} $
therefore $ -2ah+a^{2}+k^{2}=0 $
Replace (h, k) by (x, y), then $ y^{2}-2ax+a^{2}=0 $ is the required locus.
BETA
