Find the locus of Q

Question

Let $P$ be a variable point on the hyperbola $ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Then draw two perpendicular chords from $P$ to the hyperbola so as to meet at $R$ and $T$ . All the chords $RT$ are concurrent at the point $Q$ (for a given point $P$). As $P$ varies find the locus of the point $Q$?.
The question seems to be very simple but it was very hard to solve. I tried a lot to minimise the number of steps but I couldn't . The answer was found to be $$ \frac{x^2}{a^2}-\frac{y^2}{b^2}= (\frac{a^2+b^2}{a^2-b^2})^2 $$ Is there any nice geometric or a simple solution for this question ? Can any one solve it for me ? Thanks in advance.

Answer 1

I don't know if more simple solution of the problem exists. This solution is hard due to lot of calculations, but calculations are only in rational functions.

Let $P(x,y)$. Let lines $PR$ and $PT$ are the lines given by parametric equations $r(t)=(x+kt,y+t)$ and $r(t)=(x-\frac{1}{k}t,y+t)$. These lines are perpendicular and passing through $P$. Let's find point of intersection of line $PR$ with hyperbola. Putting $(x+kt,y+t)$ in hyperbola equation we can get $e(t)=\frac{(x+kt)^2}{a^2}-\frac{(y+t)^2}{b^2}-1=0$. As we know point $P$ is on the hyperbola, so $t=0$ is one of roots. Then second root can be found from linear equation $\frac{e(t)-e(0)}{t}=0$. Resulting formula is $$t=\frac{2a^2y-2b^2kx}{b^2k^2-a^2}$$

Mark this value as $t_2$. Putting $-\frac{1}{k}$ instead of $k$, we can get $t_3$. Then $R(x_2=x+kt_2,y_2=y+t_2)$, $T(x_3=x-\frac{1}{k}t_3,y_3=y+t_3)$, where $x_2, x_3, y_2, y_3$ are rational algebraic expressions in terms of $a$, $b$, $x$, $y$, $k$.

Let mark $Q(x_4,y_4)$. Line $RT$ must pass through $Q$, therefore $$x_4=\frac{x_3-x_2}{y_3-y_2}y_4+\frac{x_2 y_3-x_3 y_2}{y_3-y_2}$$

Differentiating by $k$ one can get $$0=y_4 \frac{d}{dk}\left(\frac{x_3-x_2}{y_3-y_2}\right)+\frac{d}{dk}\left(\frac{x_2 y_3-x_3 y_2}{y_3-y_2}\right)$$

Taking derivatives of rational expressions, one can get resulting formula

$$y_4=\frac{(b^2+a^2)y}{b^2-a^2}$$

Putting this formula into $$x_4=\frac{x_3-x_2}{y_3-y_2}y_4+\frac{x_2 y_3-x_3 y_2}{y_3-y_2}$$ one can get

$$x_4=-\frac{(b^2+a^2)x}{b^2-a^2}$$

Then $$\frac{x_4^2}{a^2}-\frac{y_4^2}{b^2}=\left(\frac{b^2+a^2}{b^2-a^2}\right)^2 \left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)=\left(\frac{b^2+a^2}{b^2-a^2}\right)^2$$