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
Match the following

A-q, B-p, C-r, D-s

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A-p, B-q, C-r, D-s

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A-s, B-q, C-p, D-r

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A-r, B-p, C-q, D-s

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Solution

The correct option is A A-q, B-p, C-r, D-s
a) $(y + 2)^{2} = (x - 1)$

Vertex= $(1, 2)$

b) $x^{2} - 2 x - 12 y + 13 = 0$

$(x - 1)^{2} = 12 (y - 1)$
Vertex = $(1, 1)$
$a = 3$
focus = $(1, 1 + 3)$
$= (1, 4)$
c) $(x + 2)^{2} = - 2 (y - 6)$
Vertex = $(- 2, 6)$
$a = - \frac{1}{2}$
$(y - 6) = - a$
$\Rightarrow y - 6 = \frac{1}{2}$
$y = 6 + \frac{1}{2}$
$\Rightarrow 2 y = 13$
d) $\frac{d y}{d x} = \frac{1}{2 y} = \frac{1}{2}$ $\Rightarrow$ slope of normal $= - 2$
$(y - 1) = - 2 (x - 1)$
$2 x + y = 3$

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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

List-I List-II A) Vertex of the parabola y2−x+4y+5=0p) (1,4)B) Focus of the parabola x2−2x−12y+13=0q). (1,-2) C) The directrix of the parabola x2+4x+2y−8=0r) 2y-13=0 D) Normal at (1,1) to y2=xs) 2x+y-3=0Match the following

The correct option is A A-q, B-p, C-r, D-sa)(y+2)2=(x−1)Vertex=(1,2)b)x2−2x−12y+13=0(x−1)2=12(y−1)Vertex = (1,1)a=3focus = (1,1+3)=(1,4)c)(x+2)2=−2(y−6)Vertex = (−2,6)a=−12(y−6)=−a ⇒y−6=12y=6+12⇒2y=13d)dydx=12y=12⇒ slope of normal=−2(y−1)=−2(x−1)2x+y=3

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
Match the following