1. ## Hyperbola

I'm stuck on a proof of the following:

S and S' are the foci of the hyperbola with equation x^2/a^2 - y^2/b^2 = 1.
Show that SP and S'P are equally inclinled to the tangent at any point P on the hyperbola.

I've worked out the gradients for the tangent as (b sec t) / (a tan t); the line SP as (b tan t)/(a sec t - ae); S'P as (b tan t)/(a sec t + ae). I don't know where to go from there.

2. Originally Posted by pingleby
S and S' are the foci of the hyperbola with equation x^2/a^2 - y^2/b^2 = 1.
Show that SP and S'P are equally inclined to the tangent at any point P on the hyperbola.

I've worked out the gradients for the tangent as (b sec t) / (a tan t); the line SP as (b tan t)/(a sec t - ae); S'P as (b tan t)/(a sec t + ae). I don't know where to go from there.
Two things you need to remember: (1) the trig formula $\tan(\theta-\phi) = \tfrac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi}$; (2) the formula $a^2+b^2=a^2e^2$ for the eccentricity of the hyperbola.

You know that the gradient of the tangent at P is $\tfrac{b \sec t}{a \tan t}$, which is slightly more conveniently written as $\tfrac b{a\sin t}$ (after multiplying top and bottom by cos(t)). Similarly, the gradients of SP and S'P are $\tfrac{b\sin t}{a(1\pm e\cos t)}$. Therefore, by the formula (1), the tan of the angle between the tangent at P and the line SP is $\frac{\frac b{a\sin t} - \frac{b\sin t}{a(1- e\cos t)}}{1+\frac b{a\sin t}\,\frac{b\sin t}{a(1- e\cos t)}}.$

That compound fraction simplifies (I'll skip the details) to $\frac{ab\cos t(\cos t - e)}{(a^2+b^2)\sin t - a^2e\cos t\sin t}.$

Now use (2) to replace $a^2+b^2$ in the denominator by $a^2e^2$. You will then find that the fraction simplifies further, to $-\tfrac{b\cos t}{ae\sin t}.$

That gives the tan of the angle between the tangent at P and the line SP. What about the analogous calculation for the other focus? Fortunately, you don't need to go through another long computation, because the only difference between the gradient of SP and the gradient of S'P is that e is replaced by -e. So, without any further work, you know that the tan of the angle between the tangent at P and the line S'P is $-\tfrac{b\cos t}{a(-e)\sin t}.$ In other words, it is the negative of the previous result. Therefore the two angles are equal (but with opposite sign).