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} ) $ .