Parabola Problem

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Find the vertex, axis of symmetry, focus, and directrix of each parabola. Indicate whether the vertex is a max. or min. point.

Problem 45.
x=4y^2 - 6y + 15
 

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  • #2
HallsofIvy
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Were you able to do the first 44 problems? :smile:

Here is a hint: write x= 4(y2- (3/2)y )+ 15 and then complete the square. That will tell you where the vertex is. Your text book should have a formula allowing you to find the information you need.

The directrix, in this case, is a vertical line and the axis of symmtry is a horizontal line.

Give it a try and show us what you can do.
 
  • #3
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HallsofIvy said:
Were you able to do the first 44 problems? :smile:

Here is a hint: write x= 4(y2- (3/2)y )+ 15 and then complete the square. That will tell you where the vertex is. Your text book should have a formula allowing you to find the information you need.

The directrix, in this case, is a vertical line and the axis of symmtry is a horizontal line.

Give it a try and show us what you can do.
This is what I have done:

x=(y2- (3/2)y )+ 15
= 4(y-0.75)^2 +15-.5625
= 4(y-0.75)^2+231/16

Is this right?
 
  • #4
arildno
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Apart from a lacking 4 in your first line (you remember it in the next two lines!), you get the right answer.
1.What is therefore the position of the vertex?

2. How would you approach the problem to determine the directrix and the focus?
 
  • #5
arildno
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Sorry, your expression in 1. is wrong:
You should have:
[tex]x=4(y^{2}- \frac{3}{2}y) + 15 = 4((y-0.75)^2-\frac{9}{16})+15
= [/tex]
[tex]4(y-0.75)^2+15-\frac{9}{4}=4(y-0.75)^2+\frac{51}{4}[/tex]
 
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  • #6
arildno
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Just a few hints about finding the directrix and focus:
The focus lies on the symmetry line y=0.75
Hence, let the coordinates of the focus be [tex](x_{f},0.75)[/tex]
(You must determine [tex]x_{f}[/tex])

The directrix is a vertical line; let it have the coordinates: [tex](x_{d},y),-\infty\leq{y}\leq\infty[/tex]

The parabola is all pairs (x,y) which lies in equal distance from the focus and the directrix:
[tex]\sqrt{(x-x_{d})^{2}+(y-y)^{2}}=\sqrt{(x-x_{f})^{2}+(y-0.75)^{2}}[/tex]

By squaring this equation, and comparing with your original expression, you may determine [tex]x_{f},x_{d}[/tex]
 

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