# Homework Help: The orthocentre of the triangle and a parabola

1. Apr 27, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data
The orthocentre of the triangle formed by points t1,t2, t3 on the parabola y2 = 4ax is
vertex
Origin
Focus
(1,0)

2. Relevant equations
NA

3. The attempt at a solution
The points can be taken anywhere,
So orthocentre can be formed anywhere isn't it?

2. Apr 27, 2015

### phinds

I don't understand that statement but if I am interpreting it correctly you are asking "is it true that the orthocenter can be formed anywhere?" but I still don't know exactly what you mean. What is your point? If the orthocenter CAN be formed anywhere, what does that tell you about the answer to the question?

3. Apr 27, 2015

### Raghav Gupta

Yeah, you have interpreted it correctly.
It tells me that no option satisfies for the answer.
Is it correct?

4. Apr 27, 2015

### phinds

Yes, that's how I see it too.

5. Apr 27, 2015

### Raghav Gupta

They should have specified about the points t1,t2, t3 ?

6. Apr 27, 2015

### phinds

I don't understand what you mean. They DID specify the points. The points have to be on the parabola. That IS the specification.

Now, if you mean "they should have specified something ELSE about the points, in addition to their needing to be on the parabola, so that one of the answers is correct" then I agree with you.

7. Apr 27, 2015

### HallsofIvy

I interpret this, not as "the orthocenter can be any where" but rather as "given any three points on this parabola, the orthocenter of the triangle formed by them is ...."
That is, that the orthocenter for any three points is a specific point. I haven't done any calculation but notice that only one of the points given is inside the parabola as the orthocenter would have to be.

8. Apr 27, 2015

### phinds

Nice. That may be the right way to look at it since it gives a clear answer.

9. Apr 27, 2015

### Raghav Gupta

And the clear answer is that orthocentre is definitely inside parabola but not satisfying the options.

10. Apr 27, 2015

### LCKurtz

I don't follow you here. Why do you say only one of the three points is inside the parabola? The three points are given to be on the parabola. And why is it obvious that the orthocenter would be inside the parabola? Is that even true? It the triangle has an obtuse angle its orthocenter is outside the triangle, and maybe outside the parabola.

[Edit, added] Consider the parabola $y=x^2$ and the points $(1,1),(2,4),(3,9)$. It's easy enough to calculate the coordinates of the orthocenter as $(-30,13)$ which is outside the parabola.

[Edit, added again] Never mind. These calculations were for the circumcenter. Sorry.

Last edited: Apr 27, 2015
11. Apr 27, 2015

### Raghav Gupta

I wanted to know earlier that what is the formula or how you calculate orthocentre if given 3 points of a triangle. Can you tell me?

12. Apr 27, 2015

### LCKurtz

Just find the slopes and mid points of two sides, then write the equations of their perpendicular bisectors and solve where they intersect.

13. Apr 27, 2015

### Raghav Gupta

Why the mid points, are you talking of circumcentre?
Orthocentre is the intersection of altitudes (from a vertex to opposite side).

14. Apr 27, 2015

### LCKurtz

Woops! Yes. I was thinking circumcenter in my above posts.

15. Apr 27, 2015

### phinds

I think you are missing Hallsofivy's point. If you assume that one of the answers has to be correct, then clearly then only answer possible is
Are you saying that you don't know how to use Google to look up a simple defintion?

16. Apr 28, 2015

### Raghav Gupta

I am not getting it.
vertex and origin looks same to me for this parabola.
focus and point (1,0) are inside parabola. How we can see a clear answer?

Sorry, searched finding orthocentre on google recently.

17. Apr 28, 2015

### Raghav Gupta

I think vertex is the answer but what is the reason for that?

18. Apr 28, 2015

### LCKurtz

You can draw any parabola and any three points on it and it obvious that the orthocenter of the triangle formed is not at the vertex. What is suggested by the answer choices, and what you need to prove, is that it is at the focus. Whether that is actually true or not, I haven't checked, but it seems unlikely to me.

19. Apr 28, 2015

### Raghav Gupta

How can we prove that?
(1,0) also looks a suitable answer other then focus.

20. Apr 28, 2015

### LCKurtz

Take the parabola $y=x^2$. Consider the three points $(-1,1),(1,1),(2,4)$. Sketch the parabola and the triangle. The altitude from $(2,4)$ to the horizontal side is the line $x=2$. So the orthocenter must lie on that line. It isn't inside the triangle and it can't be the focus or vertex because they lie on the line $x=0$ and it isn't $(1,0)$. Something is wrong or incomplete with this problem.

21. Apr 29, 2015

### Raghav Gupta

Yeah, but it is the @phinds guy who is creating confusion by saying something that I am not understanding properly.

22. Apr 29, 2015

### Raghav Gupta

Would I never get this problem solved?
What bad I have done @phinds that you are not replying and keeping me and @LCKurtz in confusion?

23. Apr 29, 2015

### phinds

I don't know how to state what I have stated any more clearly than I have already stated it. What is it that you want me to clarify?

24. Apr 29, 2015

### Raghav Gupta

Looking at your post #8 it looks like one of the options from problem statement might be correct . Is it so?

25. Apr 29, 2015