
#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
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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: 38,879

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

Emeritus
Sci Advisor
PF Gold
P: 16,101

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. 


Register to reply 
Related Discussions  
Parabola  Precalculus Mathematics Homework  4  
Parabola Problem  Introductory Physics Homework  1  
Verification of a parabola  Introductory Physics Homework  3  
Parabola Help Plz  Introductory Physics Homework  1  
AS advancing physics course  General Physics  2 