Conic Sections Question 40
Question: The equation of the parabola whose vertex is at (2, -1) and focus at (2, -3) is
[Kerala (Engg.) 2002]
Options:
A) $ x^{2}+4x-8y-12=0 $
B) $ x^{2}-4x+8y+12=0 $
C) $ x^{2}+8y=12 $
D) $ x^{2}-4x+12=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ VS=\sqrt{{{(2-2)}^{2}}+{{(-3+1)}^{2}}}=2 $ . From $ {{(x-h)}^{2}}=-4a(y-k) $ Parabola is, $ {{(x-2)}^{2}}=-4.2(y+1) $
therefore $ {{(x-2)}^{2}}=-8(y+1) $
therefore $ x^{2}+4-4x=-8y-8 $
therefore $ x^{2}-4x+8y+12=0. $
BETA
