Question

(12) Find the vertex, axis, focus and directrix of the parabola x2 + 4x + 2y - 7 = 0.​

Answer 1

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Let's rearrange the equation and complete the squares

y2−4y=4x

y2−4y+4=4x+4

(y−2)2=4(x+1)

Comparing this equation to

(y−b)2=2p(x−a)

p=2

The vertex is V =(a,b)=(−1,2)

The focus is F =(a+p2,b)=(0,2)

The directrix is x=a−p2

x=−1−1=−2

Answer 2

The vertex of the given parabola is = $(-2,\frac32)$

The axis of symmetry is x= -2

The focus of the parabola is =$(-2,-\frac32+(-\frac12))$$=(-2,-\frac42)=(-2,-2)$

The directrix is  y= - 1.

Step-by-step explanation:

Given equation of the parabola is

$x^2+4x+2y-7=0$

$\Rightarrow x^2+2x.2+2^2 -2^2+2y-7=0$

$\Rightarrow (x+2)^2+4+2y-7=0$

$\Rightarrow (x+2)^2=-2y+3$

$\Rightarrow (x+2)^2=-2(y-\frac32)$

$\Rightarrow (x+2)^2=4(-\frac12)(y-\frac32)$.......(1)

If equation of parabola is  $(x-h)^2=4p(y-k)$.......(2)

Then the vertex of the parabola is (h,k)

Axis of symmetry is x=h

The focus of the parabola is = (h,k+p)

Directrix is y=k-p.

Comparing the equation of equation (1) and (2)

h= -2 ,$k=\frac32$ , $p=-\frac12$

The vertex of the given parabola is = $(-2,\frac32)$

The axis of symmetry is x= -2

The focus of the parabola is =$(-2,-\frac32+(-\frac12))$$=(-2,-\frac42)=(-2,-2)$

The directrix is $y=-\frac32-(-\frac12)$ ⇒ y= - 1.