Another Ellipse question (1 Viewer)

[OLDMAN]

P is an arbitrary pt. on the ellipse and line L is the tangent to the ellipse at P.
The pts. S' and S are the foci of the ellipse. Let S" be the reflection of S across the L.

i) Prove that the focal chords through P are equally inclined.
ii) Fully describe the path of S" as P moves on the ellipse.

Morphology! : morph of q.7b HSC 2000.

 

[ND]

Ok i just had a look at q.7b of HSC 2000 and that involves euclidian geometry. Are we allowed to use a number plane for this? Part (i) would be far easier on a cartesian plane, and i wouldn't know how to describe (ii) without giving a locus.

 

[OLDMAN]

ND : good thoughtful questions.
For part i) you can say that no loss of generality in assuming the ellipse x^2/a^2+y^2/b^2=1.
For part ii) I specifically avoided the word "locus" as the only derivation of a locus eqn. required in the syllabus is with the rectangular hyperbola. A description would suffice.

 

[Harimau]

Originally posted by OLDMAN
i) Prove that the focal chords through P are equally inclined.

"Equally Inclined" Does it mean that it makes an equal angle on the X axis or something else?

 

[ND]

Re: Re: Another Ellipse question

Originally posted by Harimau


"Equally Inclined" Does it mean that it makes an equal angle on the X axis or something else?

It means that the two focal chords make equal angles with the tangent. I'll post the solution after (unless someone beats me to it), i'm lazy.

 

[ND]

Originally posted by Isaac Reyes
ND: would u happen to know a guy called Reilly Ellsmore?

Nah sorry, i don't. Where does he go to school?

 

[ND]

i) Let ellipse be x^2/a^2 + y^2/b^2 = 1 with P(acos@, bsin@).
Let A and B be the pts of intersection of the tangent at P with the perpendiculars through S and S' resp.

eqn of tangent at P:

(xcos@)/a + (ysin@)/b = 1
xbcos@ + aysin@ - ab = 0

AS = |aebcos@ - ab|/sqrt(b^2.(cos@)^2 + a^2.(sin@)^2)
= ab|ecos@ - 1|/sqrt(b^2.(cos@)^2 + a^2.(sin@)^2)

BS' = ab|ecos@ + 1|/sqrt(b^2.(cos@)^2 + a^2.(sin@)^2)

SP = a(1-ecos@) [from SP/PM = e]
S'P = a(1+ecos@)

Now sin/_SPA = AS/SP = [|aebcos@ - ab|/sqrt(b^2.(cos@)^2 + a^2.(sin@)^2)]/a(1-ecos@)
= b/sqrt(b^2.(cos@)^2 + a^2.(sin@)^2)

sin/_S'PB = BS'/S'P = [ab|ecos@ + 1|/sqrt(b^2.(cos@)^2 + a^2.(sin@)^2)]/a(1+ecos@)
= b/sqrt(b^2.(cos@)^2 + a^2.(sin@)^2)
= sin/_SPA
.'. /_SPA = /_ S'PB
.'. focal chords at P are equally inclined.

ii) i don't know...

edit: submitted before i had finished typing the soln.

 

[OLDMAN]

ND: part (ii). Hope you have drawn an ellipse, a tangent line, the two foci, and equal inclinations of the focal chords.

Can you show that S", P and S are collinear?

 

[ND]

S, P, and S'' are collinear? They don't look too collinear on the diagram in the HSC paper.

 

[OLDMAN]

Sorrry, I meant S', P and S".

 

[ND]

Ah ok.
/_SPA = /_ APS'' ('.' SP = PS" and AP is common)
/_SPA = /_ S'PB
.'. /_S'PB = /_APS"
'.' angles are vertically opp, S', P, and S" are collinear.
Is this sufficient?

 

[OLDMAN]

Yes.

 

[spice girl]

Using the property SP + S'P is a constant (=2a)
then S'P + PS" is a constant
since S'PS" is collinear, and S' is fixed
S" is a circle radius 2a about the point S'

 

[ND]

I can't believe i didn't see that. I am an idiot........