# Intersection of a tangent of a hyperbola with asymptotes

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

• Related Calculus and Beyond Homework Help News on Phys.org
mfb
Mentor
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
Homework Helper
Gold Member
$$x=\frac{b^{2} x_{1}-a^{2} y_{1}^{2}}{a y_{1} b+b^{2} x_{1}}$$ 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 mfb
Mentor
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!

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
Mentor
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
Homework Helper
Gold Member
it became very cumbersome
Only because you failed to spot that the numerators are the same, just with opposite signs.

• ElectronicTeaCup