Parabola Help Math127 2

leyva2389

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Confocal Parabola. The parabola y2=4p(x+p) has its focus at the origin and axis along the x-axis. By assigning different values to p, we obtain a family of confocal parabolas. Such families occur in the study of electricity and magnetism. Show that there are exactly two parabolas in the family that pass through a given point?

I dont know where to start. I have a photo of the picture of the confocal parabolas but for some reason it wont upload. But please help me start. I am so confused.
 
Here is the problem by the way!!!
photomath.jpg
confocal parabola. The parabola y2=4p(x+p) has its focus at the origin and axis along the x-axis. By assigning different values to p, we obtain a family of confocal parabolas. Such families occur in the study of electricity and magnetism. Show that there are exactly two parabolas in the family that pass through a given point?

I dont know where to start. I have a photo of the picture of the confocal parabolas but for some reason it wont upload. But please help me start. I am so confused.
 
Confocal Parabola. The parabola y2=4p(x+p) has its focus at the origin and axis along the x-axis. By assigning different values to p, we obtain a family of confocal parabolas. Such families occur in the study of electricity and magnetism. Show that there are exactly two parabolas in the family that pass through a given point?

I dont know where to start.
A good place to start might be with choosing a generic point, (a, b). Plug "a" in for "x" and "b" in for "y". Solve the resulting equation for "p=". Interpret the result. ;)
 
A good place to start might be with choosing a generic point, (a, b). Plug "a" in for "x" and "b" in for "y". Solve the resulting equation for "p=". Interpret the result. ;)
Or, since the problem specifically says (x1,y1)\displaystyle (x_1, y_1), use them:
y12=4p(x1p)\displaystyle y_1^2= 4p(x_1- p)
Solve for p in terms of x1\displaystyle x_1 and y1[/itex].Howmanysolutionsarethere?\displaystyle y_1[/itex]. How many solutions are there?
 
Or, since the problem specifically says (x1,y1)\displaystyle (x_1, y_1), use them:
y12=4p(x1p)\displaystyle y_1^2= 4p(x_1- p)
Solve for p in terms of x1\displaystyle x_1 and y1[/itex].Howmanysolutionsarethere?\displaystyle y_1[/itex]. How many solutions are there?
\(\displaystyle


Question. So we can pick any number value?\)
 
Question. So we can pick any number value?
Yes, you can pick any point (x1,y1). Just solve with the arbitrary numbers x1 and y1, with no special relationship between them, and solve the equation for p. Since you will have a quadratic equation for p, there will be exactly two solutions for any point (x1,y1).

BTW, in some of the postings a sign in the equation has been flipped. As I read the original equation, it should be

y12 = 4p (x1+p)\displaystyle \displaystyle y_1^2\ =\ 4p\ (x_1 + p)
 
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