Conic Sections Question 534
If the axis of a parabola is horizontal and it passes through the points (0, 0), (0, -1) and (6, 1), then its equation is y = -\frac{1}{12}(x - 6)^2 + \frac{1}{2}
Options:
A) $ y^{2}+3y-x-4=0 $
B) $ y^{2}-3y+x-4=0 $
C) $ y^{2}-3y-x-4=0 $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
Trick: There will be no constant term in a curve which passes through (0,0). So none is correct.
BETA
