The orthocentre of the triangle and a parabola

[Raghav Gupta]

Homework Statement

The orthocentre of the triangle formed by points t1,t2, t3 on the parabola y² = 4ax is
vertex
Origin
Focus
(1,0)

Homework Equations

NA

The Attempt at a Solution

The points can be taken anywhere,
So orthocentre can be formed anywhere isn't it?

 

[phinds]

So orthocentre can be formed anywhere isn't it?

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?

 

[Raghav Gupta]

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?

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

 

[phinds]

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

Yes, that's how I see it too.

 

[Raghav Gupta]

Yes, that's how I see it too.

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

 

[phinds]

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

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.

 

[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.

 

[phinds]

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.

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

 

[Raghav Gupta]

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.

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

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

 

[LCKurtz]

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.

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.

 

[Raghav Gupta]

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.

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?

 

[LCKurtz]

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?

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

 

[Raghav Gupta]

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

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

 

[LCKurtz]

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

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

 

[phinds]

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

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

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?

Are you saying that you don't know how to use Google to look up a simple defintion?

 

[Raghav Gupta]

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?

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.

 

[Raghav Gupta]

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.

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

 

[LCKurtz]

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

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.

 

[Raghav Gupta]

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.

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

 

[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.

 

[Raghav Gupta]

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

 

[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?

 

[phinds]

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?

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?

 

[Raghav Gupta]

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?

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

 

[phinds]

 

[LCKurtz]

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

There is no confusion except perhaps on your part. NONE of the answers are correct as I showed in post #20. What is confusing about that?

 

[phinds]

There is no confusion except perhaps on your part. NONE of the answers are correct as I showed in post #20. What is confusing about that?

And my solution is not correct because ... ?

 

[LCKurtz]

And my solution is not correct because ... ?

What solution? You mean that picture of the parabola in post #25? The question wasn't whether it was possible to have the orthocenter be at the (1,0). You aren't given the points are symmetric in the axis.

From your earlier posts I assumed you agree there is no correct answer to the badly stated problem, and the only person confused about this is the OP.
[Edit] corrected vertex to (1,0).

 

[LCKurtz]

What solution? You mean that picture of the parabola in post #25? The question wasn't whether it was possible to have the orthocenter be at the $(1,0)$. You aren't given the points are symmetric in the axis.

From your earlier posts I assumed you agree there is no correct answer to the badly stated problem, and the only person confused about this is the OP.

I just realized in my post #20 I used a parabola $y=x^2$ instead of $x=y^2$. But the argument still stands, none of the answers are correct.

 

[phinds]

What solution? You mean that picture of the parabola in post #25? The question wasn't whether it was possible to have the orthocenter be at the (1,0). You aren't given the points are symmetric in the axis.

From your earlier posts I assumed you agree there is no correct answer to the badly stated problem, and the only person confused about this is the OP.
[Edit] corrected vertex to (1,0).

"origin" was one of the possible answers. My solution gives the origin. Why is that not correct?

 

[phinds]

"origin" was one of the possible answers. My solution gives the origin. Why is that not correct?

EDIT: and by the way, see post #8[/QUOTE]

 

[LCKurtz]

"origin" was one of the possible answers. My solution gives the origin. Why is that not correct?

Because the question didn't ask whether it was possible for the orthocenter to be the origin. It gave three arbitrary points on the parabola and asked which answer was correct for the location of the orthocenter. As my example in post #20 shows, the orthocenter can be outside the parabola in which case none of the answers are correct. You have shown a particular collection of points having orthocenter at the origin, but there is nothing in the problem that specifies that the points must be symmetric to the axis of the parabola as yours are.

 

[phinds]

Because the question didn't ask whether it was possible for the orthocenter to be the origin. It gave three arbitrary points on the parabola and asked which answer was correct for the location of the orthocenter. As my example in post #20 shows, the orthocenter can be outside the parabola in which case none of the answers are correct. You have shown a particular collection of points having orthocenter at the origin, but there is nothing in the problem that specifies that the points must be symmetric to the axis of the parabola as yours are.

See post #8 and the one it was responding to.

 

[LCKurtz]

See post #8 and the one it was responding to.

HallsofIvy is incorrect that the orthocenter must be inside the triangle. See my post #20.
Gotta run for now.

 

[phinds]

HallsofIvy is incorrect that the orthocenter must be inside the triangle. See my post #20.
Gotta run for now.

Well, my interpretation and solution do not imply that the orthocenter MUST be inside triangle, although it certainly CAN be inside. My solution in fact has the orthocenter ON the triangle (and on the origin). The interpretation here is that they are saying there ARE three point that have one of the possible solutions as the correct answer and that's what my solution does.