Find a point on the axis of a parabola

In summary: But it would be very tedious and time-consuming to go through all of that.In summary, the author is trying to find a point on a parabola y^2 = 4ax which has the property that if a chord PQ is drawn through it, then \frac{1}{PK^2} + \frac{1}{QK^2} is the same for all positions of the chord. They are also saying that they are completely at sea when it comes to proving this. However, they think that the chord could be considered a normal at the point P and they could get the equation of the normal as y = -t_{1}x + 2at_{1} + at_{1}^
  • #1
FedEx
318
0

Homework Statement



Prove that on the axis of any parabola [tex]y^2 = 4ax[/tex] there is a certain point K which has the property that,if a chord PQ of the parabola be drawn through it ,then [tex]\frac{1}{PK^2} + \frac{1}{QK^2}[/tex] 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 [tex](at_{1}^2,2at_{1}) and (at_{2}^2,2at_{2})[/tex]

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

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

Now our aim is to show that [tex]\frac{1}{PK^2} + \frac{1}{QK^2}[/tex] is independent of [tex]t_{1} and t_2{}[/tex]. I tried and applied the distance formula but no benefit.
 
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  • #2
The theorem says that there exists such a point K. What are the conditions on K that will make [tex]\frac{1}{PK^2} + \frac{1}{QK^2}[/tex] independent of t1 and t2?
 
  • #3
HallsofIvy said:
The theorem says that there exists such a point K. What are the conditions on K that will make [tex]\frac{1}{PK^2} + \frac{1}{QK^2}[/tex] independent of t1 and t2?

Yes they are saying that there exists such a point that [tex]\frac{1}{PK^2} + \frac{1}{QK^2}[/tex] is same for all positions of that point and we have to prove this.
 
  • #4
My point was that you said:
FedEx said:
Now our aim is to show that [tex]\frac{1}{PK^2} + \frac{1}{QK^2}[/tex] is independent of [tex]t_{1} and t_2{}[/tex]. 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.
 
  • #5
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 [tex](-at_{1}t_{2},0)[/tex].

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

[tex] y = -t_{1}x + 2at_{1} + at_{1}^3 [/tex] 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.
 
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Related to Find a point on the axis of a parabola

1. How do you find the point on the axis of a parabola?

To find the point on the axis of a parabola, you can use the formula x = -b/2a, where a and b are the coefficients of the quadratic equation representing the parabola. This point is also known as the vertex of the parabola.

2. What is the significance of the point on the axis of a parabola?

The point on the axis of a parabola is important because it marks the highest or lowest point on the curve, depending on the direction of the parabola. It also helps in determining the shape and orientation of the parabola.

3. Can there be more than one point on the axis of a parabola?

No, there can only be one point on the axis of a parabola. This point is the vertex of the parabola and is located at the midpoint between the two x-intercepts.

4. How do you determine the direction of the parabola based on the point on the axis?

If the parabola opens upwards, the point on the axis will be the lowest point on the curve. If the parabola opens downwards, the point on the axis will be the highest point on the curve. This can also be determined by looking at the coefficient of the squared term in the quadratic equation.

5. Is finding the point on the axis of a parabola the same as finding the x-intercepts?

No, finding the point on the axis of a parabola is different from finding the x-intercepts. The x-intercepts are the points where the parabola intersects the x-axis and can be found by solving the quadratic equation. The point on the axis is the vertex of the parabola and can be found using the formula x = -b/2a.

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