Equation for he indicated parabola

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In summary: You should be able to see the relationship that will allow you to find the missing piece of information.I'm not sure that's the best way to do it, but that's how I got the answers. The parabola is a circle-like shape that has a focus and a directrix. The distance from the focus to the directrix is equal to the distance from any point on the parabola to the focus (this is the definition of a parabola).So, for question 2, the distance from the vertex to the focus is 2 (since the distance from the focus to the directrix is 2). Since the vertex is halfway between the focus and the directrix, the vertex is on the line y
  • #1
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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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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. :frown: 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! :smile:


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.
 

1. What is the equation for the indicated parabola?

The equation for a parabola is y = ax^2 + bx + c, where a, b, and c are constants and x is the variable.

2. How do I find the vertex of a parabola using the equation?

The vertex of a parabola can be found by using the formula x = -b/2a. Once you have the x-value of the vertex, you can plug it back into the equation to find the y-value.

3. Can you explain the significance of the coefficient "a" in the equation for a parabola?

The coefficient "a" determines the direction and width of the parabola. If a is positive, the parabola opens upward and if a is negative, the parabola opens downward. The absolute value of a also determines the width of the parabola - the larger the absolute value, the narrower the parabola.

4. How many roots does a parabola have?

A parabola can have 0, 1, or 2 roots, depending on the value of the discriminant b^2 - 4ac. If the discriminant is positive, there are 2 distinct roots. If the discriminant is 0, there is 1 real root. If the discriminant is negative, there are no real roots.

5. Can the equation for a parabola be used to model real-world situations?

Yes, the equation for a parabola can be used to model a variety of real-world situations, such as the trajectory of a projectile, the shape of a satellite dish, or the height of a roller coaster. By adjusting the values of a, b, and c, the parabola can be shifted, stretched, or compressed to fit different scenarios.

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