Find a point on the axis of a parabola

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving a parabola defined by the equation y² = 4ax. Participants are tasked with proving the existence of a specific point K on the axis of the parabola, which maintains a particular relationship involving chords drawn through it.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conditions under which the expression \(\frac{1}{PK^2} + \frac{1}{QK^2}\) remains constant for varying positions of the chord. There are attempts to derive the coordinates of point K and explore the implications of the distance formula.

Discussion Status

There is an ongoing exploration of the problem, with some participants questioning the assumptions regarding point K and its properties. While some have proposed potential coordinates for K, the proof of the stated theorem remains a point of contention and inquiry.

Contextual Notes

Participants note the need to adhere to the problem's requirements, including proving the existence of point K and its coordinates, while also grappling with the implications of the conditions set by the problem.

FedEx
Messages
318
Reaction score
0

Homework Statement



Prove that on the axis of any parabola y^2 = 4ax there is a certain point K which has the property that,if a chord PQ of the parabola be drawn through it ,then \frac{1}{PK^2} + \frac{1}{QK^2} is same for all positions of the chord.Find aslo the coordinates of the point K


Homework Equations



We can apply the parametric equations of a parabola.


The Attempt at a Solution



Let the points P and Q be (at_{1}^2,2at_{1}) and (at_{2}^2,2at_{2})

So the equation of the chord would be y(t_{1} + t_{2}) = 2x + 2at_{1}t_{2}

Hence from there we have that the points of K are (-at_{1}t_{2},0)

Now our aim is to show that \frac{1}{PK^2} + \frac{1}{QK^2} is independent of t_{1} and t_2{}. I tried and applied the distance formula but no benefit.
 
Last edited:
Physics news on Phys.org
The theorem says that there exists such a point K. What are the conditions on K that will make \frac{1}{PK^2} + \frac{1}{QK^2} independent of t1 and t2?
 
HallsofIvy said:
The theorem says that there exists such a point K. What are the conditions on K that will make \frac{1}{PK^2} + \frac{1}{QK^2} independent of t1 and t2?

Yes they are saying that there exists such a point that \frac{1}{PK^2} + \frac{1}{QK^2} is same for all positions of that point and we have to prove this.
 
My point was that you said:
FedEx said:
Now our aim is to show that \frac{1}{PK^2} + \frac{1}{QK^2} is independent of t_{1} and t_2{}. I tried and applied the distance formula but no benefit.


without any conditions on K. You are not asked to show that but rather find the single point K for which that is true.
 
Agreed. But we also have to prove that. The question says "Prove that on the axis of any parabola ...coordinates of the point K.

And let's forget about that for a minute, i have shown that the coordinates of the point k would be (-at_{1}t_{2},0).

So that is done.

But when it comes to proving,i am completely at sea.

But i think we can consider that chord to be a normal at the point P. If we do so we can get the equation of the normal as

y = -t_{1}x + 2at_{1} + at_{1}^3 and at the same time we can also consider the equation of the tangent passing through P and than we can consider a tangent at Q which will intersect the tangent at P and then we MAY get some relation.
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
15
Views
5K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
9K
  • · Replies 9 ·
Replies
9
Views
5K
Replies
3
Views
3K