The chord $QQ'$ of a hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2}=1$ is parallel to the tangent at $P$. The segments $PN$, $QM$ and $Q'M'$ are perpendiculars to an asymptote. Show that $QM \cdot Q'M' = PN^2$.
I did the following step .
Coordinates of $P,Q$ and $Q'$ are $(x_1,y_1)$ , $(x_2,y_2)$ and $(x_3,y_3)$
The asymptote equation is $\frac{x}{a} -\frac{y}{b} =0$
Using distance of Perpendicular line formula
Let $$PN = \frac{\frac{x_1}{a} -\frac{y_1}{b}}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\\ QM= \frac{\frac{x_2}{a} -\frac{y_2}{b}}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\\ Q'M'= \frac{\frac{\ x_3}{a} -\frac{\ y_3}{b}}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$$
Slope at $P$ is $\frac{b^2 x_1}{a^2 y_1}$
Slope of $QQ'$ is $\frac{y_2 - y_3}{x_2 - x_3}$
$\frac{b^2 x_1}{a^2 y_1}$=$\frac{y_2 - y_3}{x_2 - x_3}$
As $QQ'$ is parallel to Tangent at $P$ hence both slopes are equal , using above relation.
At the end I tried to compare $PN^2 = QM\cdot Q'M'$ using the "slope of $QQ'$ equals the slope of $P$" condition but getting struck.
I will present two solutions (A) and (B). I have found solution (B) 2 months after I wrote solution (A). By far, solution (B), is the simplest one.
Solution (A)
See figure 1 below.
Hyperbola $(H)$ with equation
$$\tag{A0} \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$
can be described using the classical parametrization :
$$\tag{A1} x=a \cosh(t), \ \ y=b \sinh(t).$$
Let $t,t',t''$ be the parameters corresponding to points $Q,Q',P$ resp. Let:
$$\tag{A2} d=\dfrac{1}{\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}}}.$$
The distance of point $Q$ with coordinates (A1) to the asymptote of $(H)$ (depicted in green on the figure) with equation $\frac{x}{a}-\tfrac{y}{b}=0$ is (you have done this computation):
$$\tag{A3} QM=d(\cosh(t)-\sinh(t))=d e^{-t} \ \text{; thus} \ Q'M'=d e^{-t'} \ \text{and} \ PN=d e^{-t''}.$$
(see for example http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html). Thus, what we need to prove becomes
$$d e^{-t}d e^{-t'}=(d e^{-t''})^2.$$
Said otherwise, we have to establish the (pretty) result that
$$\tag{A4}t''=\tfrac12 (t+t').$$
Let us work backwards : we are going to prove that the tangent to $(H)$ at point $P'$ of the hyperbola with parameter $\tfrac12 (t+t')$ is parallel to line segment $QQ'$. (both in blue on the picture).
This is rather easy.
$$\tag{A5}x \dfrac{1}{a} \cosh(\tfrac12 (t+t'))- y.\dfrac{1}{b} \sinh(\tfrac12 (t+t'))=1.$$
A directing vector of this tangent is thus
$$\vec{V}=\binom{\tfrac{1}{b} \sinh(\tfrac12 (t+t'))}{\tfrac{1}{a} \cosh(\tfrac12 (t+t'))}=\tfrac{1}{ab}\binom{a \sinh(\tfrac12 (t+t'))}{b \cosh(\tfrac12 (t+t'))}.$$
(I use here the fact that a straight line with equation $ux+vy+w=0$ has $\binom{v}{-u}$ as a directing vector).
$$\vec{QQ'}=2\sinh(\tfrac12 (t-t'))\binom{a\sinh(\tfrac12 (t+t'))}{b\cosh(\tfrac12 (t+t'))}.$$
Parallelism is a consequence of the fact that vectors $\vec{V}$ and $\vec{QQ'}$ are proportional.
Explanation for (A5) : the generic formula for the tangent to $(H)$ in $(x_0,y_0)$ is, using (A0) :
$$\dfrac{x x_0}{a^2}-\dfrac{y y_0}{b^2}=1.$$
Important remark: We have the interesting result $MN=NM'$. It can be easily proved using the concept of conjugate diameters.
Solution (B)
(see figure 2 below).
Let us take the following axes. Let the asymptote of hyperbola (H) on which the projection is done be the $y$ (ordinate) axis, and axes' intersection point as the origin. Thus, the other asymptote has equation $y=ax$ for a certain $a \in \mathbb{R}$, giving the following equation for (H):
$$\tag{B1}y=ax+\dfrac{b}{x} \ \ \ \text{with} \ \ \ a \in \mathbb{R}, b \in \mathbb{R^*}$$
In order to simplify our explanation, let us consider the case of equation
$$\tag{B2}h(x)=y=x-\dfrac{1}{x}$$
(The general case (B1) is an immediate generalization of (B2)).
Let $Q_k(x_k,y_k:=h(x_k)) \ \ (k=1,2)$ two points on the hyperbola with equation (B2) situated on the same branch of (H). We can assume WLOG, by the fact that $h$ is an odd function, that $x_1>0$ and $x_2>0$ .
The slope of line segment $Q_1Q_2$ is:
$$\tag{B3}s=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\left(x_2-\dfrac{1}{x_2}\right)-\left(x_1-\dfrac{1}{x_1}\right)}{x_2-x_1}=1+\dfrac{1}{x_1x_2}.$$
Let $M_k$ be the projection of $Q_k$ on the $y$ axis. As distance $Q_kM_k$ is equal to $x_k$, the issue is to show that, at point $P$ with abscissa $x_3:=\sqrt{x_1x_2}$, the tangent has a slope equal to $s$ as given in (B3).
In fact, differentiation of (B2) gives the following expression for the slope of the tangent in $x$:
$$\tag{B4}h'(x)=1+\dfrac{1}{x^2}$$ Computed for $x=x_3$, we find back at once (B3).
Remark: The slope of the tangent could have been directly obtained by taking the limit of expression (B3) when $x_1 \to x$ and $x_2 \to x$.