Intersection of a tangent of a hyperbola with asymptotes

[ElectronicTeaCup]

Summary:: Question: Show that the segment of a tangent to a hyperbola which lies between the asymptotes is bisected at the point of tangency.

Spoiler: Answer in solution booklet

Spoiler: My Attempt

From what I understand of the solution, I should be getting two values of x for the intersection that should be equivalent but with different signs. However, I get

$$x=\frac{b^{2} x_{1}-a^{2} y_{1}^{2}}{a y_{1} b+b^{2} x_{1}}$$ and $$x=-\frac{b^{2} x_{1}-a^{2} y_{1}^{2}}{a b y_{1}-b^{2} x_{1}}$$
as the x-coordinates of the intersects with the two asymptotes. But these are not equivalent, am I making a mistake somewhere?

 

[mfb]

I don't understand your approach. Did you just revert the slope on one side?

The x-values don't need to be equal in magnitude with opposite sign, but their average should be the x-value of the tangent point.

 

[ElectronicTeaCup]

I am simultaneously solving

1) The equation of the hyperbola $y-y_{1}=\frac{b^{2} x_{1}}{a^{2} y_{1}}\left(x-x_{1}\right)$ with the equation of the top asymptote $bx + ay = 0$

2) The equation of the hyperbola $y-y_{1}=\frac{b^{2} x_{1}}{a^{2} y_{1}}\left(x-x_{1}\right)$ with the equation of the bottom asymptote $bx - ay = 0$

Where $(x_{1},y_{1})$ are the coordinates of the tangent on the hyperbola.

So am I just missing the next step then? I.e to do the same with the y intercepts and plug them into the distance formula? I was assuming this would be too cumbersome—am I incorrect to think so?

 

[haruspex]

$$x=\frac{b^{2} x_{1}-a^{2} y_{1}^{2}}{a y_{1} b+b^{2} x_{1}}$$ and $$x=-\frac{b^{2} x_{1}-a^{2} y_{1}^{2}}{a b y_{1}-b^{2} x_{1}}$$

Those equations are not dimensionally consistent. Shoukd the x₁ in the numerators be squared?
Having fixed that, as @mfb indicates, try taking the average of your two x values.

 

[ElectronicTeaCup]

Oh yes, the $x_{1}$ is supposed to be squared. I tried averaging them but it became very cumbersome

Spoiler: getting average

 

[mfb]

I'm not sure which step is wrong but the expressions for the two solutions should be much simpler. WolframAlpha (I replaced x1 by c).

 

[ElectronicTeaCup]

I'm not sure which step is wrong but the expressions for the two solutions should be much simpler. WolframAlpha (I replaced x1 by c).

Oh I see! Using the answers from WolframAlpha gives me the right results. Now to find that mistake! Thank you all!

 

[ElectronicTeaCup]

There wasn't a mistake, just one more step was needed:

Is there a method to do this division, and how do you get the intuition to divide it anyway?

 

[mfb]

Ah, with the numerator fixed it's better to see. If you have a difference of squares it's often a good idea to write that as product. $a^2 y^2 - b^2 c^2 = (ay-bc)(ay+bc)$ and now the first factor is clearly in the denominator (just with an additional b).

 

[haruspex]

it became very cumbersome

Only because you failed to spot that the numerators are the same, just with opposite signs.

 

[Charles Link]

I worked through this one also, after everyone else solved it. It is a fun problem.