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PLSSSSS HELP!!! WHY IS NOBODY ANSWERING!! PARABOLAS

PLSSSSS HELP!!! WHY IS NOBODY ANSWERING!! PARABOLAS-example-1
PLSSSSS HELP!!! WHY IS NOBODY ANSWERING!! PARABOLAS-example-1
PLSSSSS HELP!!! WHY IS NOBODY ANSWERING!! PARABOLAS-example-2
PLSSSSS HELP!!! WHY IS NOBODY ANSWERING!! PARABOLAS-example-3
PLSSSSS HELP!!! WHY IS NOBODY ANSWERING!! PARABOLAS-example-4
PLSSSSS HELP!!! WHY IS NOBODY ANSWERING!! PARABOLAS-example-5
User Penz
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1 Answer

2 votes

Answer:

1.
x=0

2.
(x+5)^2=16(y-2)

3.
(x+4)^2=(2)/(3)(y-2)

4. See explanation

5.
(y-6)^2=-4(x-3)

Explanation:

1. The equation of given parabola is


(y+3)^2=4(x-1)

The vertex of this parabola is at point (1,-3), this parabola goes in positive x-direction and has the parameter
p, such that


2p=4\\ \\p=2

The directrix is the vertical line. The equation of the directrix is


x-x_0=-(p)/(2)\ [x_0\text{ is the x-coordinate of the vertex}]\\ \\x-1=-(2)/(2)\\ \\x-1=-1\\ \\x=0

2. The vertex of the parabola is at point (-5,2), the focus is at point (-5,6), this means that the vertex and the focus both lie on the vertical line x = -5. This line is the line of symmetry of the parabola. Thus, parabola goes in positive y-direction and has the equation


(x-(-5))^2=2p(y-2)\\ \\(x+5)^2=2p(y-2)

To find the parameter
p, note that the distance between vertex and focus is equal to
(p)/(2), hence,


(p)/(2)=|6-2|\\ \\(p)/(2)=4\\ \\p=8

and the equation of the parabola is


(x+5)^2=2\cdot 8\cdot (y-2)\\ \\(x+5)^2=16(y-2)

3. Given the equation


3x^2+24x-2y+52=0

Rewrite this equation as follows


3(x^2+8x)-2y+52=0\\ \\3(x^2+2\cdot 4x+4^2-4^2)=2y-52\\ \\3((x+4)^2-16)=2y-52\\ \\3(x+4)^2-48=2y-52\\ \\3(x+4)^2=2y-52+48\\ \\3(x+4)^2=2y-4\\ \\3(x+4)^2=2(y-2)\\ \\(x+4)^2=(2)/(3)(y-2)

4. Given the equation of the parabola
(y+1)^2=12(x-3)

The vertex of the parabola is at point (3,-1).

The parabola opens to the right.

Parabola's parameter is
p=(12)/(2)=6, so the focus is
(p)/(2)=(6)/(2)=3units away from the vertex.

The directrix is
p=6 units away from the focus.

The focus is at the point
(3+3,-1)=(6,-1) and the directrix has the equation


x-3=-3\\ \\x=0

5. The distance between the directrix and the focus is


p=|4-2|=2 units.

The directrix is to the right from the focus, so parabola opens to the left and has the equation


(y-y_0)^2=-2p(x-x_0),

where
(x_0,y_0) is the vertex.

The vertex is the point that is half way from the focus to the directrix, hence


x_0=(4+2)/(2)=3\\ \\y_0=6

So, the equation is


(y-6)^2=-4(x-3)

User Pshirishreddy
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