Coordinate Geometry Question 99
Question: The coordinates of the point A and B are $ (ak,0) $ and $ ( \frac{a}{k},0 ),(k=\pm 1) $ . If a point P moves so that $ PA=kPB, $ then the equation to the locus of P is
Options:
A) $ k^{2}(x^{2}+y^{2})-a^{2}=0 $
B) $ x^{2}+y^{2}-k^{2}a^{2}=0 $
C) $ x^{2}+y^{2}+a^{2}=0 $
D) $ x^{2}+y^{2}-a^{2}=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{(x-ak)}^{2}}+y^{2}=k^{2}[ {{( x-\frac{a}{k} )}^{2}}+y^{2} ] $
$ \Rightarrow (1-k^{2})(x^{2}+y^{2})-2akx+2akx+a^{2}k^{2}-a^{2}=0 $
$ \Rightarrow x^{2}+y^{2}-a^{2}=0 $ .
BETA
