Conics - Parametrics - Ellipse (1 Viewer)

[sasquatch]

Theres a question in terry lee which asks:

"P is a point on the ellipse x²/9 + y²/4 = 1. Show clearly with the aid of a diagram how to determine the position of a point Q so that their eccentric angles differ by 90*"

This is what i did:

$ = 90 + @

.:. Q[3cos(90+@), 2sin(90+@)]

That seemed a bit too simple. The answer at the back goes on about an auxilary circle and ect. Can anybody explain.

Also is the eccentric angle the angle that a line from the center of the ellipse to a point on the ellipse makes with the x-axis on the positive side of the graph (i.e. Quadrant 1 and 4 area).

Thanks.

 

[pLuvia]

Also is the eccentric angle the angle that a line from the center of the ellipse to a point on the ellipse makes with the x-axis on the positive side of the graph (i.e. Quadrant 1 and 4 area).

No

The point Q should have the same x coordinate but just lies on an auxiliary circle.
The parametric equations of P are (3cosθ,2sinθ) therefore Q has the parametic equation of (3cosθ,2sin(90+θ))

Edit: Which edition of Terry Lee do you currently have?

 

[sasquatch]

The parametric equations of P are (3cosθ,2sinθ) therefore Q has the parametic equation of (3cosθ,2sin(90+θ))

Shouldnt it be (3cos(90+θ),2sin(90+θ))?

I see in my post i put a negative sign for the first angle, but ill fix that.

The edition i have is the 5th.

Also you said that point q should have the same x-coordinate, but lie on an auxiliary circle. How would that be possible then to make an angle of 90 degrees with point P. The answer at the back says what i said. Ill copy up what they said.

"Given a point P(acosθ, bsinθ) we can mark two points P'(acosθ, asinθ) and Q'(acos(θ+90), asin(θ+90)) on the auxiliary circle such that /_P'OQ' = 90. The ordinate of Q' meets the ellipse at Q(acos(θ+90), bsin(θ+90)), thus, P, Q are points whos eccentric angles differ by 90*."

------------------------------------------------------------------------------------------------------------

I would like to ask a different question aswell relating to conics.

I got the right answer, but i just dont get the method terry lee used.

The question states

"The equation of a chord of the ellipse x² + 4y²=9 is 3x - 2y - 1 =0. Find the coordinates of its midpoint"

The method i used was to write the chord equation in terms of x and subsitute it into the equation of the ellipse. Here i solved for y, and substituted the y-values into the chord equation to find x. From here i used that midpoint formula to find the midpoint. This was a bit challenging, but terry lee finds it to be very simple.

Heres his method:

x² + 4[(3x+1)/2]² = 9
x<sup>2 + (3x-1)2 = 0
10x2 - 6x - 8 = 0

Ok this part is similar, he subsituted the solution of x for the chord equation. But here is the part i dont get completely.

Let M be the midpoint of the chord,

x_M = (1/2) * sum of roots = (1/2) * (6/10) = (3/10)
y_M = (3x_M - 1) / 2 = -1/20</sup>

<sup>the part i dont understand fully is x_M part.

I am aware that the roots of an equation resulting from subsitution of another equation mean the points of intersection.

Should i use this method? Are we expected to know this... ive never really had a proper explination, but i remember seeing it as a soultion to some question ages ago.. then for some reason it stuck. So well i dont really know what im asking, but do others use this for such purposes.

EDIT: Now i know what i dont get. Why do you use the SUM of the roots?

SECOND EDIT: the roots are the points of intersection, so (sum of roots / 2) equals the x coordinate for the midpoint! i get it now!!!

Also id like to confirm something to do with that, above.

if you subsitute together two equations and end up with a parabola. The discrimant can be used to determine if they intersect right?

/\ = 0, one root
/\ < 0, no roots
/\ > 0, two roots?

Also what happens if upon subsitution you end up with a cubic or such, is this any use?</sup>

 

[pLuvia]

Eccentric angles is the angle formed from an auxiliary circle to the centre of the ellipse + auxiliary circle and the x axis.

And therefore the x coordinate of the parametric equation for the ellipse is the same as the x coordinate of the parametric equation for the auxiliary circle

Edit to my previous post to your eccentric angle question

 

[sasquatch]

The eccentric angle of a point on an ellipse with semimajor axes of length and semiminor axes of length is the angle in the parametrization

http://mathworld.wolfram.com/EccentricAngle.html

They do not mention anything to do with a circle.

 

[Riviet]

the part i dont understand fully is x_M part.

When you find the roots of that equation, these are the two points of intersection of the ellipse and the line. Say the roots are x1 and x2. Then the midpoint's x-value would be given by (x1+x2)/2 = (sum of roots)/2
=-b/a x 1/2
=6/10 x 1/2
=3/10

Should i use this method? Are we expected to know this...

Use it if you understand it, or if you don't feel comfortable, you can find the roots and find the average of the two for the co-ordinates, although it does take longer, you will still eventually get to the answer. However, taking advantage of shortcuts can be very rewarding as you know that 4u can be alot of messy algebra that you don't want to be doing for every question.

if you subsitute together two equations and end up with a parabola. The discrimant can be used to determine if they intersect right?

/\ = 0, one root
/\ < 0, no roots
/\ > 0, two roots?

Also what happens if upon subsitution you end up with a cubic or such, is this any use?

Yes, you can use the discriminant to determine the number of intersection points, if it has any at all. If you encounter a cubic, then it simply means there must be at least 1 real root since a cubic must cross the x-axis at one point. I think there are specific cubic discriminants but they are not examined in the HSC course though. If you do come across a cubic when finding intersection points, consider general ways to solve it like first trying to factorise it through inspection or factor theorem.

I hope that helps.

Edit: I'm glad you worked out the midpoint bit yourself. Well done.

 

[sasquatch]

Thanks...

So could anybody tell me what an eccentric angle is...cuz i dont get it!!

One more thing i keep asking questions.. i need a teacher..

how do you go from:

cosa - cos b = -2sin[(a+b)/2]]cos[(a-b)/2]

and sina - sinb = 2cos[(b+a)/2]sin[(b-a)/2]

cuz i cannot do those two!

 

[STx]

To draw the eccentric angle for any point P(acosΘ,bsinΘ) on the ellipse, we first locate the point Q(acosΘ,asinΘ). Such a point belongs to the circle of radius a, centre; the origin. Noting that P and Q have the same eccentric angle for P like this: From draw a perpendicular to the x-axis to cut the circle x²+y²=a² at Q, angle QOx is the Eccentric Angle

Check a diagram for this.

 

[Riviet]

how do you go from:

cosa - cos b = -2sin[(a+b)/2]]cos[(a-b)/2]

and sina - sinb = 2cos[(b+a)/2]sin[(b-a)/2]

cuz i cannot do those two!

Firstly, it should be a before b, and there are also two more involving sina + sin b and cosa + cosb.

Consider the expansions of:
sin(A+B)=sinAcosB+cosAsinB (1)
sin(A-B)=sinAcosB-cosAsinB (2)

Subtract (2) from (1),

sin(A+B)-sin(A-B)=2cosAsinB (3)

Let X=A+B and Y=A-B

Then (X+Y)/2 = A and (X-Y)/2 = B

Substituting these into (3),

sinX-sinY=2cos[(X+Y)/2].sin[(X-Y)/2]

For sinA + sinB, add (1) and (2) and do something similar.

For cosA - cosB, consider the expansions of:
cos(A+B)=cosAcosB-sinAsinB (5)
cos(A-B)=cosAcosB+sinAsinB (6)

For the identity involving cosA - cosB, subtract (6) from (5) and for cosA + cosB, add (5) and (6) and use a similar method that I have just shown.

 

[sasquatch]

How would you know to use two of those compound angle identities. Hadnt even occured to me.... hmm..

Another thing, with your solution the angles a and b were not retained. If you see the question that i actual posted.

And also are these common identies? Because i have never seen them before.

 

[Riviet]

How would you know to use two of those compound angle identities. Hadnt even occured to me.... hmm..

I know which two to use, because by observation/inspection, I am able to see which two combinations will yield the required identity. For example, I can see that if I use sin(A+B)-sin(A-B), it will yield the identify that has sina-sinb. Try proving each of the 4 yourself using the four compound angle expansions of sin and cos.

Another thing, with your solution the angles a and b were not retained. If you see the question that i actual posted.

I simply used X and Y as any random dummy variables. Since this is an identity, you can replace X and Y with a and b respectively.

And also are these common identies? Because i have never seen them before.

We 4 uniters should be familiar with it, ie at least know how to prove the results. I don't think they're in the 2/3 unit syllabus, but we can be asked to use these results in an exam/test with a different parts to the question to guide us through.