conics help!!! (1 Viewer)

Alkenes

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hey guys

I need help with these conics questions.

For 100620121625.jpg question 8, the answer is in 100620121626.jpg and 100620121627.jpg

I did that question long ago...i don't understand the third last line in 100620121627.jpg i.e. 'but director circle of ellipse (ii)....' please explain :)

and could you do 100620121628.jpg question 10
and 100620121629.jpg question 2 (c)

:) Thanks
 

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Carrotsticks

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I'll make a start on them tomorrow unless somebody else does them by tonight =) Just PM me a reminder if I forget (I have poor memory).
 

Carrotsticks

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First question, you asked about the direction circle thing. I don't really see it that way, I see it differently and hopefully you can understand my explanation.

We know from circle geometry that if I subtend an angle from the diameter, it will always be equal to 90 degrees. Note that this is an 'if and only if' situation, so it goes both ways.

What we want to prove essentially is that PQ is the diameter and your point R(x_0,y_0) lies on the circumference of the circle.

So you did your algebra etc etc and showed (using the discriminant) that the locus of R is a circle of radius 5.

HOWEVER, the angle R is ALWAYS subtended from the chord PQ and its locus is a circle, so all we need to do now to complete the question is to prove that PQ is the diameter of this locus.

Not quite sure where the term 'director circle' comes from, but I know it to be the Fermat-Apollonius Circle. The cool thing about this circle is that suppose I were to pick any point on this circle and draw two tangents to the ellipse, they are always perpendicular to each other.

It has the same centre as the ellipse in which it inscribes AND it has a radius of .

Now comparing this with the ellipse (ii), we see that it indeed satisfies this condition, and so therefore the locus of R is the Fermat-Apollonius Circle and thus, the tangents are always perpendicular.

Now I am a bit unsure as to whether this (theorem? property?) can be examined in the HSC because I don't recall it being taught in textbooks.
 

Carrotsticks

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2(c) Ex 5.3

Recall the definition of the ellipse PS = ePM.

Well, since PQ is a latus rectum and T lies on the directrix, we can see that PM = ST so therefore PS = eST, which implies that PS/ST = e, where e is the eccentricity of the ellipse.

 

Alkenes

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Thanks a lot Carrot! :)
Could you question 10 (b) too ?

:)
2(c) Ex 5.3

Recall the definition of the ellipse PS = ePM.

Well, since PQ is a latus rectum and T lies on the directrix, we can see that PM = ST so therefore PS = eST, which implies that PS/ST = e, where e is the eccentricity of the ellipse.

 

Alkenes

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yeah that's what I am asking....i used this rule : z1= r1 cis theta 1 z2=r2 cis theta 2

so r1*r2= cis (theta 1 + theta 2)
 

Carrotsticks

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yeah that's what I am asking....i used this rule : z1= r1 cis theta 1 z2=r2 cis theta 2

so r1*r2= cis (theta 1 + theta 2)
I personally wouldn't use that principle as tempting as it may be, because that's actually where De Moivre's Theorem comes from!

I guess you could but it's a risk. You could prove the thing a lot more safely (1 line longer I guess) by expanding it and using compound angle formulae.
 

Alkenes

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oh ok thanks man...btw you haven't answered my question 10(b) yet.... do it whenever you get time :)
 

Alkanes

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Hey what textbook are these questions from? Don't think i've seen them before.
 

Alkenes

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nope it's a text book mate
And if I had a tutor, I won't be asking questions here :L
 

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