Conic Sections Question 533
Question: Vertex of the parabola $ x^{2}+4x+2y-7=0 $ is
[MP PET 1990]
Options:
A) (-2, 11/2)
B) (-2, 2)
C) (-2, 11)
D) (2, 11)
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{(x+2)}^{2}}=-2y+7+4 $
therefore $ {{(x+2)}^{2}}=-2( y-\frac{11}{2} ) $
Hence vertex is $ ( -2,\frac{11}{2} ) $ .