Some Conics Questions (2 Viewers)

[CM_Tutor]

Before I post any questions, I would like to make a couple of introductory comments about conics.

Most students don't like conics, saying it is hard. I disagree. This is (conceptual) a relatively straight-forward topic, based mostly in simple co-ordinate geometry and calculus. What makes it difficult is that the algebra is often tangled and lengthy.

So, IMO, the key to conics is to minimise (to the greatest extent possible) the algebra. Easier said than done, you might be thinking. Well, here are some suggestions.

1. Draw a diagram. Make it big. Make sure you have plenty of space to draw onto it, should that be needed.

2. Think about the question. Most questions have an obvious algebra-bash approach. You are trying to find an alternative, because there usually is one.

3. Alternatives:
(a) Look to use properties of the conic sections, like the focus-directrix definition PS = ePM.
(b) Look for recognisable geometry shapes, triangles, etc.
(c) Look for connections between those shapes - similar triangles, congruent triangles, etc
(d) Look to make diagonals into horizontals and verticals - if you have to find a distance, you want it to be horizontal or vertical so that you can avoid the distance formula.
(e) Look for other results, like intercept theorems, or trigonometry.

4. Only after you have exhausted the above may you consider an algebra-bash.

Now, here are some questions to think about - I'm looking for two types of solutions - an elegant solution (if you can find one), or an algebra bash if you can't. Have fun!

1. The point P is any point on the ellipse x² / a² + y² / b² = 1, which has foci at S and S'. Prove that PS + PS' = 2a. (Note: this is a standard problem. Anyone who does not know a really fast way to do this should learn it as soon as one of your colleagues posts it.)

2. The point P(acos@, bsin@) is a point on the ellipse x² / a² + y² / b² = 1. The tangent at P meets the x-axis at T, and the normal at P meets the x-axis at N. Draw a diagram to represent this information, and show that
(PT / PN)² = tan²@ / (1 - e²).

3(a). ABC is a triangle and X is a point on BC. Prove that AX bisects angle BAC if and only if AB:AC = BX:CX, using the sine rule or otherwise.

(b) P(asec@, btan@) is any point on the hyperbola x² / a² - y² / b² = 1, which is not located on the x-axis. The tangent at P meets the x-axis at T, and S and S' are the foci of the hyperbola.

(i) Show that T has coordinates (acos@, 0)
(ii) Prove that PS and PS' are inclined to the tangent at P at equal angles.

4. P(acos@, bsin@) and Q(acos#, bsin#) are two distinct points on the ellipse x² / a² + y² / b² = 1, and the chord PQ subtends a right angle at A(a, 0). Find an expression for tan(@ / 2) * tan(# / 2) in terms of a and b only.

5. P(asec@, btan@) is a point on the hyperbola x² / a² - y² / b² = 1, and tan@ <> 0. The tangent at P meets the x- and y- axes at X and Y, respectively, and meets the asymptotes at K and L.

(a) Find the equation of the tangent at P, and hence find the coordinates of X and Y.

(b) Find the value of
(i) PX / XY, and
(ii) PX / PY

(c) Find the value of PK / PL

 

[grimreaper]

for q1:
if M is a point on the directrix closest to P,
SP\PM=e
SP=PM x e
Similarly for S'P:
S'P = PM' x e
SP + S'P = e(PM + P'M)
Distance from M to M' = 2a/e since they are on the line x = +/- a/e so
SP + S'P = e x 2a/e = 2a

 

[grimreaper]

Hmm well my mid course is tomorrow and I cant do 2 or 3 (I only tried 2 using distance formula, it got real ugly) so I guess I'm pretty screwed for conics lol, luckily I'm not bad at polynomials and complex no.'s...

 

[DcM]

keke q1 was from cambridge..

for q2..is there a shortcut
or do u have to find PT and PN and expand out?

 

[grimreaper]

Originally posted by DcM
keke q1 was from cambridge..

for q2..is there a shortcut
or do u have to find PT and PN and expand out?

I tried that for q2 and... well... lets just say I gave up when i saw what it all expanded out to using the distance formula

 

[DcM]

the way that im approaching q2 so far:

PNT is a triangle
i made a line thru P that cuts X axis perpendicularly at C
then i found NC^2 + PC^2 = PN^2
and similarly CT^2 + PC^2 = PT^2
and then i'm solving PT^2/PN^2

but i get this huge weird algebra..

CM am i on the rite track?

 

[DcM]

argh..ill do this tmw or some other time..
eyes getting tired..

oh btw for q3 how do u represent a variable X on a triangle?
do u give it co ordinates?

or can u prove that they r similar some other way?

 

[CM_Tutor]

Q1 is a standard question - it would be found (I expect) in every textbook, and it's an example I use to demonstrate that doing algebra is a pain in the posterior when you don't need to.

Re question 2, the obvious ('algebra-bash') approach would be to find the coordinates of N and T and use the distance formula. I wouldn't recommend this approach. DcM, your approach is using Pythagoras' theorem to prove the distance formula - I like the initiative, and the initial observation (PNT is a triangle) - but you are also headed for an algebra bash.

A good tip to ask with conics questions is what you actually care about - Do you really care what PN and PT actually are?. All I'll say (as a hint) is that I didn't find either of these distances...

Oh, and as for 3(a), it might help to note that it's a problem in plane geometry, not conics...

 

[Faera]

3a.

Let angle XAB = angle XAC = x
let angle AXC = a,
therefore angle AXB = 180 - a

in triangle ACX:
sinx / CX = sina / AC ....1

in triangle ABX:
sinx / BX = sin(180-a) / AB
sinx / BX = sina / AB ....2

1 / 2 :

BX/CX = AC/BC
which means that
BX:CX = AC:BC
as required.


b.
Tangent at p:
(can't be bothered typing out the derivation...)
[xsec@ / a] + [ytan@ / b] = 1
but when the tangent crosses the x axis, y = 0
therefore x = a/sec@ = acos@
therefore T has co-ordinates (acos@, 0)

c.

PS and PS' will be inclined at the same angle to PT if PS/PS' = ST/S'T (from question a).

ST/S'T = (ae - acos@)/(acos@ - ae)
= (e - cos@)/(cos@ - e)

PS/PS' = ..... ( i used the distance formula here- i see no other way of doing it, uless there are similar triangles somewhere that I missed, but yeah... and i CANNOT be stuffed typing the distance formula out- so...)
= (e - cos@)/(cos@ - e)

Therefore PSS' = ST/S'T, and thus Angle SPT = angle S'PT
therefore the lines PS and PS' are inclined to the tangent at P at equal angles.

 

[CM_Tutor]

Note that 3(a) said "if and only if". This means that you need to prove that

if AX bisects angle BAC then AB:AC = BX:CX

AND

if AB:AC = BX:CX then AX bisects angle BAC.

Faera, you have proven the first of these, but not the second. However, your answer to (c) relies on the second.

Also, your proof of (c) has a problem, as you have written that

ST/S'T = (ae - acos@)/(acos@ - ae) = (e - cos@)/(cos@ - e)

Unfortunately, (e - cos@) / (cos@ - e) = -(cos@ - e) / (cos@ - e) = -1.
Thus, ST/S'T = -1

There is a problem with the ratio of two distances being negative...

 

[DcM]

hey for q2..

does tan ( tan ^-1 ((a/b) tan @)) = (a/b) tan @
like can u just cancel out the tan..?

 

[DcM]

for 3c

instead of using PS/PS'
i used PM/PM' which is the same..since PS = e.PM
and it shud b easier cos its a horizontal line..

 

[DcM]

q4 was just using the m1 * m2 = -1...
answer : - b^2/a^2

for q5
a) was normal subbing into tangent

b) i) i used the similar triangle thing
so PX/XY became PC/OY where C a point where the vertical line from P cuts the X axis..so the distances became easy to calculate as its just vertical lines

ii) did the same as above..by drawing a line from P horizontally to cut the Y axis so its similiar triangle and easy to calculate..

c) is there a quick way?
so far ive made PK/PL = KB/AL where
B: x coord = K x coord
y coord = P y coord
and

A: x coord = L x coord
y coord = P y coord


so yea...thats as far as i got..

CM, pls tell me if there are quicker ways to do the questions that i just did..

thanks =)

 

[picaresque]

Q2.
tangent at P: xcos@/a + ysin@/b = 1
at T, y=0, hence x=a/cos@

From RHS
tan^2@/(1-e^2)
= tan^2@/(b^2/a^2)
=[(sin@/cos@)*(a/b)]^2
=[asin@/bcos@]^2

hence,
Prove: (PT/PN)^2=[asin@/bcos@]^2

RTP: PT/PN=asin@/bcos@

In triangle PCT & triangle PNT (let C be where the perpendicular from P cuts the x-axis)
angle NPT = angle PCT (both right angles)
angle PTC is common

therefore, triangle PCT ||| triangle PNT

hence,
PT/TC=PN/PC
=> PT/PN=TC/PC
(TC=(a/cos@)-acos@)
(PC=bsin@)
... (I'm skipping a few boring steps)
PT/PN=asin@/bcos@

therefore:
(PT/PN)^2=[tan^2@/1-e^2]

 

[DcM]

mmm...

thats not the way i did it..but i dunno if i did correctly..

let angle PNT = @
PT/PN = tan @

@ = tan^-1 (a tan@/b) since mPN = a tan@/b

:. PT/PN = a tan@/b
:. (PT/PN)^2 = a^2 tan^2@/ a^2 (1 - e^2)
= tan^2@/(1 - e^2)

...
thats y i asked if for tan( tan^-1 x)..can u just cancel the tan and tan^-1

 

[picaresque]

Originally posted by DcM

let angle PNT = @

I didn't think this was possible, since P is (acos@, bsin@) - @ would be the angle subtended at the origin. The only case where angle PNT would be @ is when your normal at P passes through the origin. Doesn't that mean your solution doesn't apply as P moves around the ellipse?

Correct me if I'm wrong. I'm still trying to come to terms with this topic. >.<

 

[CM_Tutor]

Picaresque, your solution to q2 is certainly better than finding both coordinates and using the distance formula, but DcM's way is the shortcut I was looking for (with 1 minor alteration).

DcM, nice spotting on the method for 2 - trigonometry on the right angled triangle PNT is the best approach, but I have one little problem with your answer.

You start with "let angle PNT = @" - this is a problem, as @ is already defined, and it isn't angle PNT. You needed to use another (undefined) variable.

I'll come back to q 5 later ...

 

[George W. Bush]

I was thinking about trig with the right angled triangle as well, but how do you relate the angle PN makes with the x-axis to theta? I thought about sine rule with the triangle POT but it didn't really work out

 

[CM_Tutor]

Originally posted by George W. Bush
I was thinking about trig with the right angled triangle as well, but how do you relate the angle PN makes with the x-axis to theta? I thought about sine rule with the triangle POT but it didn't really work out

Don't relate the angle PN makes to the x-axis to theta, it isn't necessary, and it's a definitional mistake on DcM's part.

Let angle PNT be alpha.

Now, tan alpha = PT / PN, using the right angled triangle.
But, tan alpha = m_PN, using 2u coordinate geometry.
So, instead of proving that (PT / PN)² = tan²@ / (1 - e²), it is sufficient to prove that m_PN² = tan²@ / (1 - e²)

Now, m_PN = -1 / m_PT = a² * bsin@ /( b² * acos@) = atan@ / b
So, m_PN² = a²tan²@ / b² = a²tan²@ / [a²(1 - e²)] = tan²@ / (1 - e²), as required.

 

[George W. Bush]

CM: Is there a easy way to solve the following question without extensive algebra?

The points P and Q lie on an ellipse (standard eqn). The tangents and P and Q intersect at point T. The midpoint of PQ is the point M. Prove that OM extended (where O is origin) passes through the point T.