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Equation for he indicated parabola 
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#1
Apr2503, 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 2xy=0 3.vertex(3,0), focus (0,1) please tell me the steps how to find this 3 parabola equation... 


#2
Apr2503, 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.



#3
Apr2503, 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 


#4
Apr2503, 10:38 PM

P: n/a

Equation for he indicated parabola



#5
Apr2603, 09:09 AM

Math
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PF Gold
P: 39,363

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
Apr2603, 09:12 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,363

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_{0})^{2}+ y_{0}. 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 y^{2} and xy. That's the result of the rotation of axes. 


#7
Apr2603, 10:28 AM

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P: 16,092

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: d^{2} = (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 ) d^{2} = (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: x^{2}  2xy + y^{2}  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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