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Ellipses and eccentricity

Hi, can someone explain when I would need to use the property PS = ePM

It crops up in part c of the question but I didn’t know how to do it. e is root(3) /2

I took a general point P (x, y) and set the distance of it from the foci as 2 IMG_8373.jpeg

Reply 1

Original post
by subbhy
Hi, can someone explain when I would need to use the property PS = ePM
It crops up in part c of the question but I didn’t know how to do it. e is root(3) /2
I took a general point P (x, y) and set the distance of it from the foci as 2 IMG_8373.jpeg

Naively youd use tthe equation (part a) and pythagoras/circle from part b) to solve a couple of quadratic equations?

Reply 2

Original post
by mqb2766
Naively youd use tthe equation (part a) and pythagoras/circle from part b) to solve a couple of quadratic equations?


Yes, that’s what I started to do

Reply 3

Original post
by subbhy
Yes, that’s what I started to do

It would help to see what youve done / what youve had a problem with when you post the question, but if you started doing that, what was the problem, why do you think you should use PS = ePM ...?

Reply 4

Original post
by mqb2766
It would help to see what youve done / what youve had a problem with when you post the question, but if you started doing that, what was the problem, why do you think you should use PS = ePM ...?

I did it this way IMG_9091.jpeg


But the mark scheme uses PS = ePM so I was wondering why that works / when I need to use this / what cues are in the question that suggest I can use this

Reply 5

Original post
by subbhy
I did it this way IMG_9091.jpegBut the mark scheme uses PS = ePM so I was wondering why that works / when I need to use this / what cues are in the question that suggest I can use this

From looking at the mark scheme (it would have been helpful to post it), the alternative method is what you did. The main method uses the focus-directrix property so for any point on an ellipse, the ratio of its distance to a focus and its distance to its corresponding directrix is constant (equal to the eccentricity). I suppose as the question is based on specifying the ellipse in terms of the directrices, and as it avoids solving quadratic simultaneous equations, it is preferable. Similarly, 5 marks is a bit stingy to do simultaneous quadratic equations (sort of twice and evaluate y). One to remember if youre given an ellipse and directrices.

Reply 6

Original post
by mqb2766
From looking at the mark scheme (it would have been helpful to post it), the alternative method is what you did. The main method uses the focus-directrix property so for any point on an ellipse, the ratio of its distance to a focus and its distance to its corresponding directrix is constant (equal to the eccentricity). I suppose as the question is based on specifying the ellipse in terms of the directrices, and as it avoids solving quadratic simultaneous equations, it is preferable. Similarly, 5 marks is a bit stingy to do simultaneous quadratic equations (sort of twice and evaluate y). One to remember if youre given an ellipse and directrices.

Thanks. I also did a similar thing for this Q but again they used PS = ePM

Why does it work here IMG_9121.jpegIMG_9124.jpeg

Reply 7

Original post
by subbhy
Thanks. I also did a similar thing for this Q but again they used PS = ePM
Why does it work here IMG_9121.jpegIMG_9124.jpeg

Id have thought its fairly self explanatory, especially with a sketch. PM and PM' are the distances from the directrices and are equal to both PS and PS' (both divided by e). So their sum is constant (distance between directrices) ...
(edited 9 months ago)

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