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Equation for he indicated parabola

by newton1
Tags: equation, parabola
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newton1
#1
Apr25-03, 09:50 PM
P: 152
find aan equation for he indicated parabola
1.focus(1,2), directrix x+y+1=0
2.vertex(2,0), directrix 2x-y=0
3.vertex(3,0), focus (0,1)
please tell me the steps how to find this 3 parabola equation...
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climbhi
#2
Apr25-03, 10:12 PM
P: n/a
Sounds like a homework question, you know the rules, show us what you did where you got stuck and then we help you from there.
newton1
#3
Apr25-03, 10:16 PM
P: 152
actually i read the book myself
because my teacher skip this topic
so i not so understand
eh...the parobola had the property that
d(P,F)=d(P,l)
for every point P(x,y)
and focus F, directrix l

climbhi
#4
Apr25-03, 10:38 PM
P: n/a
Equation for he indicated parabola

Originally posted by newton1
actually i read the book myself
Ahh well that's good to hear, but I can't help you out, I'm not sure about how to handle the non horizontal directrixes in 1 & 2, sorry! Perhaps someone else knows how to deal with this.
HallsofIvy
#5
Apr26-03, 09:09 AM
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Here's how I would do the first problem: The given directrix is a line at 45 degrees to the x axis. Set up new,x' y', coordinates: x= -x'+ y', y= x'+ y'. We can add the two equations to get 2y'= x+y or y'= (1/2)x+ (1/2)y. Subtracting the first equation from the second, we get 2x'= x- y so x'= (1/2)x- (1/2)y.

The reason for that choice is that sin(45)= cos(45)= [sqrt](2)/2. A rotation would be x= -([sqrt](2)/)x' +([sqrt](2)/2)y',
y= ([sqrt](2)/2)x'+([sqrt](2)/2)y'. I don't like writing all those squareroots again and again so i "stretched it also". The point is that the line x+ y+ 1= 0 become (-x'+ y')+ (x'+ y')+1= 2y'= 1 so in these coordinates, the directrix is y'= -1/2, a horizontal line.

Since the focus is at (x,y)= (1,2), in this new coordinate system it is at x'= (1/2)1- (1/2)2= -1/2 and y'= (1/2)1+(1/2)2= 3/2. The vertex of a parabola is always half way between the focus and directrix so the vertex is at (x', y')= (-1/2, 1/2).

Since the directrix is now horizontal, the axis is vertical and this parabola has equation y'= (4/c)(x'-x<sub>0</sub>)<sup>2</sup>+ y<sub>0</sub>. In this case that is (c= 3/2- 1/2= 1)
y'= (1/4)(x'+1/2)<sup>2</sup>+ 1/2.

Now go back to x and y: y'= (1/2)x+ (1/2)y and x'= (1/2)x- (1/2)y so (1/2)x- (1/2)y= (1/4)((1/2)x-(1/2)y)<sup>2</sup>+ 1/2.

Notice that this will involve both y<sup>2</sup> and xy. That's the result of the rotation of axes.
HallsofIvy
#6
Apr26-03, 09:12 AM
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Here's how I would do the first problem: The given directrix is a line at 45 degrees to the x axis. Set up new,x' y', coordinates: x= -x'+ y', y= x'+ y'. We can add the two equations to get 2y'= x+y or y'= (1/2)x+ (1/2)y. Subtracting the first equation from the second, we get 2x'= x- y so x'= (1/2)x- (1/2)y.

The reason for that choice is that sin(45)= cos(45)= [sqrt](2)/2. A rotation would be x= -([sqrt](2)/)x' +([sqrt](2)/2)y',
y= ([sqrt](2)/2)x'+([sqrt](2)/2)y'. I don't like writing all those squareroots again and again so i "stretched it also". The point is that the line x+ y+ 1= 0 become (-x'+ y')+ (x'+ y')+1= 2y'= 1 so in these coordinates, the directrix is y'= -1/2, a horizontal line.

Since the focus is at (x,y)= (1,2), in this new coordinate system it is at x'= (1/2)1- (1/2)2= -1/2 and y'= (1/2)1+(1/2)2= 3/2. The vertex of a parabola is always half way between the focus and directrix so the vertex is at (x', y')= (-1/2, 1/2).

Since the directrix is now horizontal, the axis is vertical and this parabola has equation y'= (4/c)(x'-x0)2+ y0. In this case that is (c= 3/2- 1/2= 1)
y'= (1/4)(x'+1/2)2+ 1/2.

Now go back to x and y: y'= (1/2)x+ (1/2)y and x'= (1/2)x- (1/2)y so (1/2)x- (1/2)y= (1/4)((1/2)x-(1/2)y)2+ 1/2.

Notice that this will involve both y2 and xy. That's the result of the rotation of axes.
Hurkyl
#7
Apr26-03, 10:28 AM
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I think it would be easier to appeal to the focus/directix definition of a parabola for problem 1 (the definition newton quoted).

The square distance from the point (x, y) to the focus is:

d2 = (x - 1)2 + (y - 2)2


To find the square distance from the directix to (x, y) we look up the point to line distance formula in the book! I can't find it in my CRC handbook so I'll compute it with cross products:

Choose the point A = (0, -1) which lies on the directix.
The unit vector v = (1/sqrt(2), -1/sqrt(2)) points along the directix.
The distance from P = (x, y) to the directix is then:
d = |(P - A) * v| = |(x, y + 1) * (1, -1) / sqrt(2)|
= |x * (-1) - (y + 1) * 1| / sqrt(2)

squaring gives: (I can flip the sign because it's inside ||)

d2 = (x + y + 1)2 / 2

And then the equation of the parabola is:

(x - 1)2 + (y - 2)2 = (x + y + 1)2 / 2

Simplifying yields:

x2 - 2xy + y2 - 6x - 10y + 9 = 0


Halls approach will work too, though there is at least 1 typo in his post. I guess it's a matter of taste which approach you use, so I suggest you do it both ways to be familiar with them both!


Incidentally, the above can be done entirely as vector equations:

Let F be the focus
Let A be any point on the directix
Let v be a unit vector that points along the directix
Let P be (x, y)
. means dot product
* means cross product

Then

(P - F).(P - F) = |(P - A) * v|2

I just recalled another formula for distance from a point to a line that's a little simpler to manipulate, allowing us to replace the absoulte value of a cross product with dot products. It gives:

(P - F).(P - F) = (P - A).(P - A) - ((P - A).v)2




For #2 and #3, you can use geometric arguments to locate the missing piece of information... think about the relationship between the focus, vertex, directerix, and the line through the focus and vertex.


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