Conic-Sections Question 576
Question: Equation of the parabola whose vertex is $ (-3,-2), $ axis is horizontal and which passes through the point $ (1,2) $ is
Options:
A) $ y^{2}+4y+4x-8=0 $
B) $ y^{2}+4y-4x+8=0 $
C) $ y^{2}+4y-4x-8=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Since the axis is horizontal an vertex is $ (-3,-2), $
$ \therefore $ The equation of the parabola must be of the form $ {{(y+2)}^{2}}=4a(x+3) $ It passes through $ (1,2), $ so $ 16=16a $ .i.e., a = 1. Hence, the equation of the required parabola is $ {{(y+2)}^{2}}=4(x+3) $ or $ y^{2}+4y-4x-8=0 $
BETA
