Conic Sections Question 154
Question: If the normal to $ y^{2}=12x $ at (3, 6) meets the parabola again in (27, -18) and the circle on the normal chord as diameter is
[Kurukshetra CEE 1998]
Options:
A) $ x^{2}+y^{2}+30x+12y-27=0 $
B) $ x^{2}+y^{2}+30x+12y+27=0 $
C) $ x^{2}+y^{2}-30x-12y-27=0 $
D) $ x^{2}+y^{2}-30x+12y-27=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
According to question, equation of circle with points (3, 6) and (27, -18) on diameter will be $ (x-3)(x-27)+(y-6)(y+18)=0 $
$ x^{2}+y^{2}-30x+12y-27=0 $ .
BETA
