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Problem 45.

x=4y^2 - 6y + 15

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- Thread starter mustang
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Problem 45.

x=4y^2 - 6y + 15

- #2

HallsofIvy

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Here is a hint: write x= 4(y

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.

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HallsofIvy said:

Here is a hint: write x= 4(y^{2}- (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=(y

= 4(y-0.75)^2 +15-.5625

= 4(y-0.75)^2+231/16

Is this right?

- #4

arildno

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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]

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]

Last edited:

- #6

arildno

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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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