# Loci of centres in Ellipse

1. Aug 14, 2015

### gianeshwar

Dear Friends!
If we have a variable triangle formed by joining a point of ellipse and its two focii shown in figure,how to find locii of various centres.Can you please suggest some reference to find these .I have tried by using simple method for incentre locus which I left as it goes long. Though I am still thinking.

2. Aug 14, 2015

### Staff: Mentor

First off, locus is singular and loci (one 'i') is plural, not "locii."

For your ellipse, the two foci are at $(\pm c, 0)$, where $c = \sqrt{a^2 - b^2}$). Possibly you could write an equation for the locus of the various centers as the point P on the ellipse moves around.

3. Aug 15, 2015

### HallsofIvy

Staff Emeritus
The center of such a triangle is the point whose coordinates are the means of the corresponding coordinates of the vertices.

Here, you have a ellipse given by $\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$. Assuming that a> b, the two foci have coordinates $(-\sqrt{a^2- b^2}, 0)$ and $(\sqrt{a^2- b^2}, 0)$. You can write a point on that ellipse as $P= (a cos(t), b sin(t)$ for parameter t between 0 and $2\pi$. "Average" those coordinates.

(That turns out to be remarkably simple and gives a very easy answer!)

Last edited: Aug 15, 2015
4. Aug 16, 2015

### gianeshwar

Thanks Mark44 and Hallsoflvy!

5. Aug 19, 2015

### gianeshwar

#### Attached Files:

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6. Aug 19, 2015

### gianeshwar

Dear friends ! In the solution of finding locus of incentre ,I am getting parabola as locus which does not seem to be by physical picturization.

7. Aug 19, 2015

### HallsofIvy

Staff Emeritus
Well, what did you do when you followed our suggestions? What is the "average" of the three points $(-\sqrt{a^2- b^2}, 0)$, $(\sqrt{a^2- b^2}, 0)$, and $(a cos(t), b sin(t))$? It is NOT a parabola! Please show exactly what you did.

8. Aug 19, 2015

### gianeshwar

Thanks Hallsfloy! I believe you are telling to find centroid locus which I am showing now. It is an ellipse which is fine.The first locus which I had calculated was of incentre in attached picture earliar.

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9. Aug 19, 2015

### HallsofIvy

Staff Emeritus
Yes, that's right. As I said originally, "That turns out to be remarkably simple and gives a very easy answer!" The y coordinates of the first two points are 0 and their x coordinates cancel so the equation of the locus is just the original ellipse, divided by 3. It is the ellipse with major and minor axes 1/3 the length of the original ellipse.

10. Aug 19, 2015

### gianeshwar

Thanks! Now my doubt in #6 is still left!

11. Aug 19, 2015

### HallsofIvy

Staff Emeritus
What "doubt" do you mean? The difficulty you stated in #6 was that you got a parabola which, of course, can't be right because a parabola would have to go outside the original ellipse. But in #8 you say you got an ellipse as the answer.

12. Aug 19, 2015

### Raffaele

Incenter is the intersection point of bisector lines of the angles of the triangle
Circumcenter is the intersection point of the sides perpendicular bisectors
Ortocenter is the intersection point of the height
Centroid is the intersection point of the medians

The average of the coordinates gives the centroid.
The other points are less trivial to find
Incenter can be found using the formula here http://www.mathopenref.com/coordincenter.html

Circumcenter P has coordinates
$$x_P=\frac{{x_A}^2 {x_B}-{x_A}^2 {x_C}-{x_A} {x_B}^2+{x_A} {x_C}^2-{x_A} {y_B}^2+{x_A} {y_C}^2+{x_B}^2 {x_C}-{x_B} {x_C}^2+{x_B} {y_A}^2-{x_B} {y_C}^2-{x_C} {y_A}^2+{x_C} {y_B}^2}{2 (-{x_A} {y_B}+{x_A} {y_C}+{x_B} {y_A}-{x_B} {y_C}-{x_C} {y_A}+{x_C} {y_B})}$$

$$y_P= \frac{{x_A}^2 ({x_C}-{x_B})+{x_A} \left({x_B}^2-{x_C}^2+{y_B}^2-{y_C}^2\right)-{x_B}^2 {x_C}+{x_B} \left({x_C}^2-{y_A}^2+{y_C}^2\right)+{x_C} ({y_A}-{y_B}) ({y_A}+{y_B})}{2 ({x_A} ({y_B}-{y_C})+{x_B} ({y_C}-{y_A})+{x_C} ({y_A}-{y_B}))}$$

Orthocenter O has coordinates

$$x_O=\frac{{x_A} ({x_B} ({y_B}-{y_A})+{x_C} ({y_A}-{y_C}))-({y_B}-{y_C}) ({x_B} {x_C}+({y_A}-{y_B}) ({y_A}-{y_C}))}{{x_A} ({y_B}-{y_C})+{x_B} ({y_C}-{y_A})+{x_C} ({y_A}-{y_B})}$$

$$y_O=\frac{{y_C} (-{x_A} {y_A}+{x_B} {y_B}+{x_C} ({y_A}-{y_B}))+({x_A}-{x_B}) (({x_A}-{x_C}) ({x_B}-{x_C})+{y_A} {y_B})}{{x_A} ({y_B}-{y_C})+{x_B} ({y_C}-{y_A})+{x_C} ({y_A}-{y_B})}$$

It can be quite hard to work it out...

Graphically it can be seen as in the following picture

Last edited: Aug 19, 2015
13. Aug 19, 2015

### gianeshwar

I am referring to my first picture where I have evaluated locus of incentre.

14. Aug 19, 2015

### gianeshwar

Picture in #5

15. Aug 19, 2015

### gianeshwar

Thanks Raffaelel! I have calculated incentre locus in attached image in #5.Need to get it verified.

Last edited: Aug 19, 2015