4Units Problem-Solving section (1 Viewer)

[victorling]

i have tried searching for questions
but almost all have been answered...

 

[Wohzazz]

Ok, let's test your skill.

The tangent at a point P of the ellipse x*2/a*2 + y*2/b*2 cuts the x axis at T and the perpendicular PN is drawn to the x axis. Prove that OT.ON=a*2

if you could, take some 'accepted shortcuts'

 

[victorling]

Originally posted by Wohzazz
Ok, let's test your skill.

The tangent at a point P of the ellipse x*2/a*2 + y*2/b*2 cuts the x axis at T and the perpendicular PN is drawn to the x axis. Prove that OT.ON=a*2

if you could, take some 'accepted shortcuts'

ok

 

[victorling]

Originally posted by victorling
ok

solution:

Let x = acos@, y = bsin@ for point P,

At P, dx/d@ = -acos@, dy/d@ = bcos@

therefore, dy/dx at P is dy/d@ * (1/(dx/d@))

= bcos@/(-asin@)

= b/-a(tan@)

Therefore the equation of the tangent at P to the ellipse is:

y - bsin@ = (-b/a(tan@)) * (x-acos@)

sub y = 0 into this equation gives u the coordinates of T,
which is x = asin@tan@ + acos@

= a((sin@)^2 + (cos@)^2) /cos@


= a/cos@

Therefore T is (a/cos@, 0)

We know PN is perpendicular to the x-axis, therefore the x-coordinate of N would be equal to that of P's, which is also acos@

therefore N is (acos@, 0)


ON*OT = (acos@-0)* ((a/cos@)-0)

= a^2

 

[Wohzazz]

Damn that was easy. Heh, guess i'm crap. I didn't convert tan@ to sin@/cos@ and just left it. I knew it was easy. Thanks victorling.

 

[spice girl]

u can solve the case for the circle x^2 + y^2 = a^2 using similar triangles, and then squish it into an ellipse. because you're not changing the x-co-ords of any point during this transformation, the value OT*ON is still preserved.

 

[victorling]

Originally posted by Wohzazz
Damn that was easy. Heh, guess i'm crap. I didn't convert tan@ to sin@/cos@ and just left it. I knew it was easy. Thanks victorling.

no problem!

 

[KeypadSDM]

Originally posted by spice girl
u can solve the case for the circle x^2 + y^2 = a^2 using similar triangles, and then squish it into an ellipse. because you're not changing the x-co-ords of any point during this transformation, the value OT*ON is still preserved.

Just HAD to do it differently didn't you?

 

[freaking_out]

Originally posted by KeypadSDM
Just HAD to do it differently didn't you?

yep, and innovatingly as well.

 

[Wohzazz]

Originally posted by spice girl
u can solve the case for the circle x^2 + y^2 = a^2 using similar triangles, and then squish it into an ellipse. because you're not changing the x-co-ords of any point during this transformation, the value OT*ON is still preserved.

i don't really get what you saying, can you lay it out more thoroughly, if it is shorter, it will help heaps
........else i'll have to do it the long 'standard' way

 

[victorling]

Originally posted by Wohzazz
i don't really get what you saying, can you lay it out more thoroughly, if it is shorter, it will help heaps
........else i'll have to do it the long 'standard' way

no worry, that 's an alternative solution
the most important thing now is that u know how to approach it

 

[turtle_2468]

Umm, spice girl is spicing it up
Probably the other one is easier to understand and apply generally... unless you have an affinity for geometric transformations, ha ha...

 

[victorling]

o
the maths god, turtle

 

[freaking_out]

Originally posted by turtle_2468
Umm, spice girl is spicing it up
Probably the other one is easier to understand and apply generally... unless you have an affinity for geometric transformations, ha ha...

true...i'm wondering if u have your own method as well?

 

[KeypadSDM]

Originally posted by freaking_out
true...i'm wondering if u have your own method as well?

I'm wondering how drbuchernan's going to relate it back to the assumption that the Reinmann hypothesis is correct.

 

[Wohzazz]

Originally posted by victorling
no worry, that 's an alternative solution
the most important thing now is that u know how to approach it

was the question answered? what's the alternative method?

 

[turtle_2468]

Probably the best alternative method is spice girl's... Basically, "stretch" the plane in the direction of the y-axis. As you do that, the x-values don't change, so OT and ON remain constant. As a result, as the minor axis of the ellipse lies on the y-axis, you can "stretch" the plane such that the ellipse becomes a circle, making things much much nicer.

I'm sure you could coord bash it as well

 

[victorling]

Originally posted by turtle_2468
Probably the best alternative method is spice girl's... Basically, "stretch" the plane in the direction of the y-axis. As you do that, the x-values don't change, so OT and ON remain constant. As a result, as the minor axis of the ellipse lies on the y-axis, you can "stretch" the plane such that the ellipse becomes a circle, making things much much nicer.

I'm sure you could coord bash it as well

but answering this way in an exam, with the need of explaining the above thing, is risky for wohazz
my method is the traditional one and actually does not require the brain...i think it is more practical, considering the tension during the exam....
but anyway, who does not prefer an fast alternative!

 

[spice girl]

well Wohzazz asked for 'accepted shortcuts' so...

anyway, my answer was just a shortcut that u can only do with 5% of conics questions, and its unlikely u can make a repeat performance of it in the HSC

 

[Wohzazz]

Originally posted by spice girl
well Wohzazz asked for 'accepted shortcuts' so...

anyway, my answer was just a shortcut that u can only do with 5% of conics questions, and its unlikely u can make a repeat performance of it in the HSC

i'll stick with the traditional method
turtle_2468's method is too hard to explain in exams
i remember one of my teacher put 'interesting method' as a comment on one of the question in a different way....wasn't too sure if it was meant as a compliment or not