The vertex of the parabola $y^{2} + 4 x - 2 y + 3 = 0$ is.

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Solution

$y^{2} + 4 x - 2 y + 3 = 0$
$(y - 1)^{2} + 4 x - 1 + 3 = 0$
$(y - 1)^{2} = - (4 x + 2)$
$(y - 1)^{2} = - 4 (x + \frac{1}{2})$
vertex $= (\frac{- 1}{2}, 1)$ .

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x^{2} + 4 x + 2 y - 7 = 0

y^{2} - 4 x - 2 y - 3 = 0

List-I	List-II
A) Vertex of the parabola $y^{2} - x + 4 y + 5 = 0$	p) (1,4)
B) Focus of the parabola $x^{2} - 2 x - 12 y + 13 = 0$	q). (1,-2)
C) The directrix of the parabola $x^{2} + 4 x + 2 y - 8 = 0$	r) 2y-13=0
D) Normal at (1,1) to $y^{2} = x$	s) 2x+y-3=0

y^{2} + 2 y + x = 0

y^{2} + 4 x + 4 y - 3 = 0

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