Find the vertex, focus, and directrix of the parabola and sketch its graph.? 1. x + y^2 = 0 2. 4x - y^2-2y-33 = 0 3. x^2 - 2x + 8y + 9 =0 Please help me out with these problems, I have no idea how...

Find the vertex, focus, and directrix of the parabola and sketch its graph.? 

1. x + y^2 = 0

2. 4x - y^2-2y-33 = 0

3. x^2 - 2x + 8y + 9 =0

Please help me out with these problems, I have no idea how to do these.

Asked on by amorales63

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embizze | High School Teacher | (Level 1) Educator Emeritus

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(1) The standard form for a parabola opening left/right is `y^2=4px` if the vertex is at the origin. The directrix is the line x=p, and the focus is at (p,0)

Given `x+y^2=0` we can rewrite as `y^2=-x` or in standard form `y^2=4(-1/4)x` .(Set 4p=-1 to solve for p)

This is a parabola with vertex at the origin. Sine the coefficient is negative it opens to the left. The directrix will be p units to the right of the vertex, while the focus will be p units to the left of the vertex, along the line of symmetry.

The line of symmetry is the x-axis so we have vertex(0,0), focus `(-1/4,0)` , and directrix `x=1/4`

The graph:

(2) The standard form for a parabola opening horizontally with vertex (h,k) is `(y-k)^2=4p(x-h)`

Given `4x-y^2-2y-33=0` we isolate the y terms:

`y^2+2y=4x-33`   complete the square on the left-hand side

`(y+1)^2=4x-32`

`(y+1)^2=4(x-8)`

This is in standard form: the vertex is at (8,-1), the parabola opens right since p=1>0, the focus is 1 unit right of the vertex along the axis of symmetry (y=-1), and the directrix is the line x=7.

Thus the vertex (8,-1), focus (9,-1), directrix x=7

(3) The standard form of a parabola opening vertically with vertex(h,k) is `(x-h)^2=4p(y-k)`

Given `x^2-2x+8y+9=0` we isolate the x terms:

`x^2-2x=-8y-9`  complete the square on the left hand side:

`(x-1)^2=-8y-8`

`(x-1)^2=4(-2)(y+1)`

Thus p=-2 and we have:

The vertex is at (1,-1); the parabola opens down; the focus is at (1,-3) and the directrix is the line y=1

 

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