Help with Conic: Hyperbola's chord of contact

Question

please help with this proof.

"Show that the tangents at the endpoints of a focal chord of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ meet on the corresponding directrix."

This is a homework question with two part where the first part is to prove the converse of the above statement (namely prove that the chord of contact from a point on directrix is a focal chord)

For the first part I was able to utilize the following equation for chord of contact from a point $P(x_0, y_0)$

\begin{align} \frac{x_0 x}{a^2} - \frac{y_0 y}{b^2} = 1 \end{align}

Then substitute $x_0$ for $\pm \frac{a}{e} $ (x coordinate of directrix) and then show that $S(\pm ae, 0) $ (coordinate of focuses) satisfy the equation.

But it seem I could not use the same method to show the second part.

Please help. Thanks.

Answer 1

Use parametric form

$P_0:(x_0=a\sec\alpha,y_0=b\tan\alpha)$ and $P_1:(x_1=a\sec\beta,y_1=b\tan\beta)$

Find the abscissa$(h)$ of the intersection of the tangents at $P_0,P_1$ to be $a\dfrac{\sin(\alpha-\beta)}{\sin\alpha-\sin\beta}$

Now apply the condition of the collinearity of $P_0,P_1,S$ to get $e =\pm\dfrac{\sin\alpha-\sin\beta}{\sin(\alpha-\beta)}$

So, $h\cdot e=\pm a$ as $\sin\alpha-\sin\beta,\sin(\alpha-\beta)\ne0$

Answer 2

This likely isn’t quite the way that whoever wrote the problem had in mind, but both directions are straightforward to prove using pole/polar relations: the polar line of a point external to the hyperbola contains the chord of contact.

Write the equation of the hyperbola in matrix form as $$\mathbf x^TC\mathbf x=\begin{bmatrix}x&y&1\end{bmatrix}\begin{bmatrix}\frac1{a^2}&0&0\\0&-\frac1{b^2}&0\\0&0&-1\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}=0.$$ Using homogeneous coordinates, a point $\mathbf p$ lies on the line $\mathbf l$ iff $\mathbf p^T\mathbf l=0$. (This is just the point-normal form of the equation of a line written a bit differently.) The polar line of a point $\mathbf p$ is $C\mathbf p$ and the pole of a line $\mathbf l$ is $C^{-1}\mathbf l$. The former is the same as the formula that you have for the chord of contact for the point $(x_0,y_0)$.

So, for the first part, we need to verify that each focus lies on the polar line of a point $(\pm a/e,y)$ on the corresponding directrix: $$\begin{bmatrix}\pm a e&0&1\end{bmatrix}\begin{bmatrix}\frac1{a^2}&0&0\\0&-\frac1{b^2}&0\\0&0&-1\end{bmatrix}\begin{bmatrix}\pm\frac ae\\y\\1\end{bmatrix}=\begin{bmatrix}\pm a e&0&1\end{bmatrix}\begin{bmatrix}\pm\frac1{ae}\\-\frac y{b^2}\\-1\end{bmatrix}=0.\qquad\checkmark$$

For the other direction, we take an arbitrary line $\mathbf l=[\lambda:\mu:\mp \lambda ae]$ through the focus $(\pm ae,0)$ and compute its pole. The resulting point must lie on the corresponding directrix, i.e., on the line $[1:0:\mp a/e]$. $$\begin{bmatrix}1&0&\mp\frac ae\end{bmatrix}\begin{bmatrix}a^2&0&0\\0&-b^2&0\\0&0&-1\end{bmatrix}\begin{bmatrix}\lambda\\\mu\\\mp\lambda ae\end{bmatrix}=\begin{bmatrix}1&0&\mp\frac ae\end{bmatrix}\begin{bmatrix}\lambda a^2\\-\mu b^2\\\pm\lambda ae\end{bmatrix}=0.\qquad\checkmark$$

In fact, a focus and its corresponding directrix form a pole-polar pair: $$\begin{bmatrix}\frac1{a^2}&0&0\\0&-\frac1{b^2}&0\\0&0&-1\end{bmatrix}\begin{bmatrix}\pm ae\\0\\1\end{bmatrix}=\begin{bmatrix}\pm\frac ea\\0\\-1\end{bmatrix}.$$ The homogeneous vector on the right-hand side is equivalent to the Cartesian equation $\pm\frac eax-1=0$, or $x=\pm\frac ae$.

Answer 3

Recall that a focus is the pole of the corresponding directrix. The tangents at the endpoints of a focal chords meets at its pole which belongs to the polar of the focus, hence it belongs to the directrix.