Help with proof

[Function]

Hello everyone

 

In math class, we've proven something that had to do something with tha tangent of a hyperbola:

the tangent point P is the middle point of [QR] with Q the intersection of the tangent and one assymptote, and R the intersaction of the tangent with the other assymptote.

 

After we've calculated the values for:

 

[math]Q\left( \frac{a\cdot\cos{t}}{1-\sin{t}};\frac{b\cdot\cos{t}}{1-\sin{t}} \right)[/math]

 

[math]R\left( \frac{a\cdot\cos{t}}{1+\sin{t}};\frac{-b\cdot\cos{t}}{1+\sin{t}} \right)[/math]

 

[math]P\left( \frac{a}{\cos{t}};b\cdot\tan{t} \right)[/math]

 

we started expressing that P is middle of [QR], and getting an equality.

 

However: I claimed this could've gone a bit more easy (my math teacher asked me therefore to present her a proof):

 

P is the middle of [QR]

[math]\Leftrightarrow Q [/math] and [math]R[/math] are point symmetrical over P.

 

[math]\Leftrightarrow |x_q-x_p|=|x_p-x_r|[/math]

[math]\Leftrightarrow \left|\frac{a\cdot\cos{t}}{1-\sin{t}}-\frac{a}{\cos{t}}\right|=\left|\frac{a}{\cos{t}}-\frac{a\cdot\cos{t}}{1+\sin{t}}\right|[/math].

 

But what now? I can't just scrap the "absolute value"-signs..

 

Is there a way to proof that equation?

 

Thanks!

 

-Function

Edited September 24, 2013 by Function

[Function]

(Of course, I don't simply want to quadrate both sides)

[studiot]

Not quite sure what you are after.

 

You seem to be proving your theorem by coordinate geometry, using a trigonometric paramaterisation.

 

Are you sure you don't mean x= a cosh t and y = b sinh t ? (as opposed to cos and sin)

 

 

It is also possible to prove this by classical geometry.

 

Proposition:

 

"The portion of any tangent to a hyperbola intercepted between the asymptotes is bisected at its point of contact."

 

Let the tangent at any point P of a hyperbola cut the asymptotes in L and L'.

 

Join P to the focus, s and draw SK parallel to one asynptote CL produced.

 

Draw perpendiculars LM and L'M' to SP from L and L'

 

Then the asymptote CL produced is a tangent, whose point of contact is at infinity. Therefore LS makes equal angles with PS and SK.

 

Thus the perpendicular from L to SP is equal to the perpendicular from L to SK and therefore equal to the perpendicular t from S to CL produced and is equal to BC.

 

Thus LM = BC = L'M'

 

sinc ethe perpendiculars are equal

 

LP = PL'

 

It is also possible using an algebraic parametrisation.

 

Here it is for a rectagular hyperbola ( the X and Y axes are then the asymptotes) xy = c² : x = ct; y = ct^-1

 

Tangent is y - ct^-1 = -t^-2(x-ct)

 

When y = o ie at Q

 

-ct^-1 = -t^-2(x-ct)

c = xt^-1 - c

 

x = 2ct, which is twice the x coordinate of a point that also satisfies the hyperbola (x=ct) ie the point of contact or tangency ie the x coordinate of P

 

When x = 0 ie the point R

 

y-ct^-1 = -t^-2(0-ct)

 

y = 2ct^-1

 

so the y coordinate is also twice the y coordinate of the point of tangency, p, satisfying the parabola.

Edited September 25, 2013 by studiot

[Function]

Well, I just want to know if the last equation in my first post can be proven, in the form of, for example, |a| = |-a|, which is correct, and thus it is proven.

A quick sketch of the tangent with tangent point P and intersections with asymptotes Q and R:

 

 

To be proven: P is the middle of [QR] with Q and R given (in my first message).

I wanted to prove this, other than by using the formula to get the distance |PQ| and |PR| or using the formula of calculating the mdidle of a segment ((x1+x2)/2;(y1+y2)/2).

Therefore, I reasoned this:

In order for P to be the middle of [QR], 2 requirements must be met:

 

|xp-xr| = |xq-xp| and |yr+yp| = |yq-yp|. (I use the point symmetrical property: Q and R are point symmetrical over P)

 

When I fill in the coördinates of R, Q and P, you get the last equation of my first message.

Can that equation be proven, so I get something in the form of |a| = |-a| at the end, and thus I get something which proves that P is the middle of [QR]?

 

Thanks.

 

(Sorry for my lack of mathematical English, btw... I live in Belgium, and we don't really handle mathematical/scientifical English)

Edited September 26, 2013 by Function

[Function]

Ok, new reasoning: you may assume that it is true that P is the middle of [QR], and the given coördinates for P, Q and R are right.

 

Here's my new reasoning; I only want to know if it's plausible as proof

-------

 

For the expression:

 

[math]|x_q-x_p|=|x_p-x_r|[/math]

 

[math]\Leftrightarrow \left| \frac{a\cdot \cos{t}}{1-\sin{t}} - \frac{a}{\cos{t}} \right| = \left| \frac{a}{\cos{t}} - \frac{a\cdot\cos{t}}{1+\sin{t}} \right|[/math]

 

There are 2 solutions:

 

1) [math]\frac{a\cos{t}}{1-sin{t}} - \frac{a}{\cos{t}}=\frac{a}{\cos{t}}-\frac{a\cos{t}}{1+\sin{t}}[/math]

 

[math]\Leftrightarrow \frac{\cos^2{t}}{(1-\sin{t})\cos{t}} - \frac{1-\sin{t}}{\cos{t}(1-\sin{t}} = \frac{1+\sin{t}-cos^2{t}}{\cos{t}(1+\sin{t})}[/math]

 

[math]\Leftrightarrow \frac{1-\sin{t}+\cos^2{t}}{\cos{t}(1-\sin{t})} = \frac{1+\sin{t}-\cos^2{t}}{\cos{t}(1+\sin{t})}[/math]

 

[math]\Leftrightarrow 1-\sin{t}+\cos^2{t}+\sin{t}-\sin^2{t}+\sin{t}\cos^2{t}=1+\sin{t}-\cos^2{t}-\sin{t}-\sin^2{t}+\sin{t}\cos^2{t}[/math]

 

[math]\Leftrightarrow \cos^2{t}=-\cos^2{t}[/math]

 

[math]\Leftrightarrow \cos{t}=0[/math]

 

[math]\Leftrightarrow t=\frac{\pi}{2}+k\cdot\pi[/math]

 

2) [math]\frac{a\cdot \cos{t}}{1-\sin{t}} - \frac{a}{\cos{t}} = \frac{-a}{\cos{t}} + \frac{a\cdot\cos{t}}{1+\sin{t}}[/math]

 

[math]\Leftrightarrow 1-\sin{t}=1+\sin{t}[/math]

 

[math]\Leftrightarrow \sin{t}=0[/math]

 

[math]\Leftrightarrow t=k\cdot\pi[/math]

 

For the expression:

 

[math]|y_r+y_p|=|y_q-y_p|[/math]

 

[math]\Leftrightarrow \left| \frac{-b\cos{t}}{1+\sin{t}} + b\tan{t} \right| = \left| \frac{b\cos{t}}{1-\sin{t}}-b\tan{t} \right|[/math]

 

There should've been 2 solutions, were it not that [math]\frac{\pi}{2}+k\cdot\pi[/math] is also a solution for [math]t[/math].

 

[math]\Rightarrow \frac{\pi}{2}+k\cdot\pi[/math] is not a solution for [math]t[/math].

 

[math]\Rightarrow[/math] if the expression [math]|y_r+y_p|=|y_q-y_p|[/math] has [math]k\cdot\pi[/math] as a solution for [math]t[/math], what had to be proven, is proven.

 

[math]\Leftrightarrow \left| \frac{-b\cos{t}}{1+\sin{t}} + b\tan{t} \right| = \left| \frac{b\cos{t}}{1-\sin{t}}-b\tan{t} \right|[/math]

 

[math]\frac{-b\cos{t}}{1+\sin{t}} + b\tan{t}=\frac{-b\cos{t}}{1-\sin{t}}+b\tan{t}[/math] is a solution for the equality.

 

[math]\Leftrightarrow 1+\sin{t}=1-\sin{t}[/math]

 

[math]\Leftrightarrow \sin{t}=0[/math]

 

[math]\Leftrightarrow t=k\cdot\pi[/math]

 

Both the expressions [math]|y_r+y_p|=|y_q-y_p|[/math] and [math]|x_q-x_p|=|x_p-x_r|[/math] both have [math]k\cdot\pi[/math] as a solution for [math]t[/math]. The expressions are true.

 

I'm actually afraid that isn't a proof that the expressions are true... They only express that [math]t[/math] must be [math]k\cdot\pi[/math], which isn't the case. I'm sure my reasoning is true; where's my miscalculation?

Edited November 2, 2013 by Function

[studiot]

Why can you not simply take the average, rather than mess about with moduli?

 

If P is halfway between R and Q then

 

[math]{X_P} = \frac{1}{2}\left( {{X_R} + {X_Q}} \right)[/math]

and the same for Y_P

So substituting your values and doing a bit of manipulation

 

[math] = \frac{1}{2}\left( {\frac{{a\cos t}}{{1 - \sin t}} + \frac{{a\cos t}}{{1 + \sin t}}} \right)[/math]

 

[math] = \frac{1}{2}\left( {\frac{{a\cos t + a\cos t\sin t + a\cos t - a\cos t\sin t}}{{\left( {1 + \sin t} \right)\left( {1 - \sin t} \right)}}} \right)[/math]

[math] = \frac{1}{2}\frac{{2a\cos t}}{{1 - {{\sin }^2}t}}[/math]

[math] = \frac{{a\cos t}}{{{{\cos }^2}t}}[/math]

[math] = \frac{a}{{\cos t}}[/math]
as required

I will leave you to show the same for the Y values.

Edited November 2, 2013 by studiot

[Function]

Why can you not simply take the average, rather than mess about with moduli?

 

Because I wanted to prove it another way. What you give me is exactly that what I didn't want to see, for we had already proven it in class

[studiot]

Well I did offer you 2 alternative proofs originally, but you ignored them both.

 

If you must pursue your analysis you should carry your trigonometric manipulations through to the end.

 

[math]\frac{{1 - \sin t + {{\cos }^2}t}}{{\cos t(1 - \sin t)}}[/math]

[math] = \frac{{1 - \sin t + {{\sin }^2}t - 1}}{{\cos t(1 - \sin t)}}[/math]

[math] = \frac{{\sin t}}{{\cos t}}[/math]

[math] = \tan t[/math]

 

and

[math]\frac{{1 + \sin t - {{\cos }^2}t}}{{\cos t(1 - \sin t)}}[/math]

[math] = \frac{{{{\sin }^2}t + \sin t}}{{\cos t(1 + \sin t)}}[/math]

[math] = \frac{{\sin t}}{{\cos t}}[/math]

[math] = \tan t[/math]

[Function]

Now that is what I wanted Thanks.