Conics: open to all (1 Viewer)

[mercury]

P is any point on an ellipse with diameter QR. PR meets the tangent at Q in T. Show that the tangents at P bisects QT.

 

[ND]

When you say "diameter", do you mean major axis?

edit: nevermind, when the major axis is QR, this doesn't hold true, do you need to make Q(acos@, bsin@) and R[acos(@+pi), bsin(@+pi)] ? anyway i'll try after.

 

[ND]

Nope, that didn't work either, maybe i'm misinterpreting the question.

 

[McLake]

[Assuming diameter = major axis]

Let ellipses be of the "standard" form (I can't be bothered typing it out).

So Q(acosx, bsinx)

tangent Q: [x*cos(x)/a] + [y*sin(x)/b] = 1
x*cos^2(x) + y*sin^2(x) = 1

PR:

.....

I'll get back to you later, I havn't done this in a while ...

 

[mercury]

the diameter is not the major axis.

 

[McLake]

Originally posted by mercury
the diameter is not the major axis.

Explain ...

 

[ND]

Originally posted by McLake
[Assuming diameter = major axis]

Let ellipses be of the "standard" form (I can't be bothered typing it out).

So Q(acosx, bsinx)

tangent Q: [x*cos(x)/a] + [y*sin(x)/b] = 1
x*cos^2(x) + y*sin^2(x) = 1

PR:

.....

I'll get back to you later, I havn't done this in a while ...

If QR is the major axis it doesn't work out, the ordinate of the midpoint of QT is -(b*sin@)/(cos@-1), and the ordinate of PR at the abscissa of the MP of QT is b(cos@+1)/sin@.

Mercury, what exactly is the diameter of an ellipse?

 

[mercury]

let's say P and Q are two points on the ellipse, if PQ goes through the centre of the ellipse, then it is considered a diameter
an ellipse has infinite number of diameters

 

[ND]

I tryed making Q(acos@, bsin@) and R[acos(@+pi), bsin(@+pi)], which fits your description, but i still wasn't able to show it. Do you know how to do it?

 

[mercury]

hmm, i am attempting the more geometrical way of doing the question... and it's looking better than all those messy algebra I had before.

searching for answer....

 

[RIZAL]

Hey all,

I'm at uni now and I don't have time to work this one out.............BUT my oldman has always been interested in your problems/posts mercury so i'll email him and get him to do it.

So sit tight kids, solution coming up soon!

 

[OLDMAN]

An interesting problem but sadly lacking in leadups. In real 4unit exam, this would would be softened by a few leading questions eg.

if Q(x1,y1) and P(x2,y2) are pts in an ellipse, prove that the the intersection S of the tangents is
a^2(y2-y1)/(x1y2-x2y1),b^2(x2-x1)/(y1x2-y2x1)

if QR is a diameter, prove that R is (-x1,-y1)

Prove that OS (O being the Origin)is parallel to PR : This miight be the "hardest" part of the problem. Proof - Gradient of OS is
-(b^2(x2-x1))/(a^2(y2-y1)) and gradient of PR is
(y2+y1)/(x2+x1) Equalize and cross multiply to prove true.

These two gradients are equal since Q and R satisfy the ellipse equation x^2b^2+y^2b^2=a^2b^2. (Hard to write equations with these keys)

Since OS is parallel to PR, these lines must also split QP equally as they have QR. Proven.

 

[ND]

Originally posted by OLDMAN
An interesting problem but sadly lacking in leadups. In real 4unit exam, this would would be softened by a few leading questions eg.

if Q(x1,y1) and P(x2,y2) are pts in an ellipse, prove that the the intersection S of the tangents is
a^2(y2-y1)/(x1y2-x2y1),b^2(x2-x1)/(y1x2-y2x1)

if QR is a diameter, prove that R is (-x1,-y1)

Prove that OS (O being the Origin)is parallel to PR : This miight be the "hardest" part of the problem. Proof - Gradient of OS is
-(b^2(x2-x1))/(a^2(y2-y1)) and gradient of PR is
(y2+y1)/(x2+x1) Equalize and cross multiply to prove true.

These two gradients are equal since Q and R satisfy the ellipse equation x^2b^2+y^2b^2=a^2b^2. (Hard to write equations with these keys)

Since OS is parallel to PR, these lines must also split QP equally as they have QR. Proven.

Nice, do you teach maths?

 

[OLDMAN]

ND,

Thanks, impressed with you deciphering solution through the limited maths notation .

Not a teacher, but a maths tutor.

 

[spice girl]

Here's a funky solution:

Prove the case is true for a circle x^2 + y^2 = a^2 (which shouldn't be very hard at all, and only involves euclidean geometry). Here it is below, anyway:

In a circle, angle QPR is 90', so you'll find triangle QPT is right angled at P.

If QT and the tangent at P intersect at X, then QX = PX (tangents equal)

Now angle XPT = pi/2 - angle XPQ
= pi/2 - angle XQP (XPQ is isosceles from above)
= angle XTP

So TXP is isosceles and TX = PX

thus TX = XQ as required (for a circle)

Now, consider the transformation (affine transformation, for the professors and absolute nerds) that does this to every point on the x-y plane...

X = x
Y = (b/a)y

...to give an ellipse on the new X-Y plane.

Now lines that touch the circle at one point should still touch the new ellipse at one point and thus remain tangents.

Also, ratio of intervals on the same line should remain the same.

Thus TX = PX still holds.

 

[ND]

Originally posted by spice girl
Here's a funky solution:

Prove the case is true for a circle x^2 + y^2 = a^2 (which shouldn't be very hard at all, and only involves euclidean geometry). Here it is below, anyway:

In a circle, angle QPR is 90', so you'll find triangle QPT is right angled at P.

If QT and the tangent at P intersect at X, then QX = PX (tangents equal)

Now angle XPT = pi/2 - angle XPQ
= pi/2 - angle XQP (XPQ is isosceles from above)
= angle XTP

So TXP is isosceles and TX = PX

thus TX = XQ as required (for a circle)

Now, consider the transformation (affine transformation, for the professors and absolute nerds) that does this to every point on the x-y plane...

X = x
Y = (b/a)y

...to give an ellipse on the new X-Y plane.

Now lines that touch the circle at one point should still touch the new ellipse at one point and thus remain tangents.

Also, ratio of intervals on the same line should remain the same.

Thus TX = PX still holds.

Ooooh, spice girl is back.

 

[OLDMAN]

I have resisted making the following comment. I am glad Spice Girl used the word "funky". Indeed it is. Indeed it is an affine transformation, and invoking the fact that this transforomation preserves collinearity and ratios of distances (eg. and particularly re. this problem, the midpoint of a line segment remains midpoint after transformation). Argument is true, and I must confess I like it.

However a caveat for the year 12 student. The "chief" examiner in all his wisdom should pass it as a solution. However, the mechatronic markers (bureacratic mathematicians) working off their guidelines and tables of possible solutions may not. It is not enough to know your stuff, but one must show that you play by their rules.

 

[underthesun]

If you pay to have your paper recalculated, does the chief get to do it? Last year my maths teacher, which happens to be a marker told us how the marking tables etc works, and the fact that if there's a solution that is outside the table (The table itself is quite complex), they have to discuss together? It seems that marking is not that easy

p.s teacher is greg bamford. by any chance know him?

 

[OLDMAN]

Hope I didn't offend any marker by my "mechatronic marker" remark. Just wanted to stress that markers operate under very strict guidelines, and tremendous time constraint. Better to not plan on testing the system and to know the syllabus as best as you can.

 

[spice girl]

u're right; i wouldn't dare to write that solution in a real exam

cos this was the first time i used that method to solve a conics problem and i was surprised it worked too