1
$\begingroup$

If $PN$ is the perpendicular from a point on a rectangular hyperbola $x^2-y^2=a^2$ on any of its asymptotes, then find the locus of the midpoints of $PN.$

On drawing a rough sketch of the graph, I find that it is a rectangular hyperbola. Let a point $P$ on the hyperbola be ($\mathrm{asec\theta} , \mathrm{atan\theta}$). Asymptotes are $x^2=y^2$, that is, $y=x$ and $y=-x$.

So the distance $r$ of $P$ from $y=x$ is $\frac{a|sec\theta- tan\theta|}{\sqrt{2}}$

Now using the formula $x_1= h +rcos\theta \mathrm{\:and\:} y_1= k+sin\theta$, where $sin\theta=\frac{1}{\sqrt{2}}$ and $cos\theta=-\frac{1}{\sqrt{2}}$,

I can find the midpoint in terms of original coordinates, and hence put it in the original equation to get the locus. But this is a long method and I'm looking for an elegant way.

$\endgroup$

2 Answers 2

4
$\begingroup$

Denote our asymptotes here as $L_{+}: x - y = 0$ (the diagonal line) and $L_{-}: x + y = 0$ the off-diagonal line. We have \begin{align} x^2 - y^2 &= a^2 \\ \implies \frac{x + y}{\sqrt{2}} \frac{x - y}{\sqrt{2}} &= \frac{ a^2 }2 \\ \implies D(P, L_{-}) \cdot D(P, L_{+}) &= \frac{ a^2 }2 \tag*{Eq.(1)} \end{align} where per the setting $P$ is a point on the hyperbola, and $D(\text{point, line})$ is the perpendicular distance between the point and the line.

This is one of the basic properties of any hyperbola that the product of the distances to the asymptotes are constant.

Denote the midpoint of $PN$ as $M$, then by definition $|MN| = \frac12 |PN|$.

Consider the situation where $N$ is on $L_{+}$, which means

$$D(M, L_{+})=|MN|= \frac12 |PN| = \frac12 D(P, L_{+})\tag*{Eq.(2)} $$

Now, for our rectangular hyperbola, the distance to the other asymptote remain the same $$D(M, L_{-}) = D(P, L_{-}) \tag*{Eq.(3)}$$ because the asymptotes are perpendicular $L_{+} \perp L_{-}$.

Denote the coordinates of $M$ as $(x_1, y_1)$ and use them in the last step of rewriting Eq.(1):

\begin{align} \text{take Eq.(2) into Eq.(1)}& & \implies && D(P, L_{-}) \cdot 2 D(M, L_{-}) &= \frac{ a^2 }2 \\ \text{from Eq.(3)}& & \implies && D(M, L_{-}) \cdot D(M, L_{-}) &= \frac{ a^2 }4 \\ && \implies && \frac{ x_1 + y_1}{ \sqrt{2} } \cdot \frac{ x_1 - y_1}{ \sqrt{2} } &= \frac{ a^2 }4 \\ \end{align} This gives us the hyperbola $\displaystyle (x_1)^2 - (y_1)^2 = \frac{a^2}2$ with the same asymptotes just with a different "constant".


$\endgroup$
4
  • $\begingroup$ Nice.........+1 $\endgroup$
    – nonuser
    Commented Feb 6, 2019 at 13:54
  • $\begingroup$ So is it assumed that the locus must be another hyperbola, with the axes of it remaining same? $\endgroup$
    – user638473
    Commented Feb 6, 2019 at 14:26
  • $\begingroup$ I don't know why you use the word "assumed". It is shown (solved) to be such a hyperbola. $\endgroup$ Commented Feb 6, 2019 at 15:06
  • $\begingroup$ Oh, I get it. Thanks! $\endgroup$
    – user638473
    Commented Feb 6, 2019 at 16:50
3
$\begingroup$

This problem gets easier when you rotate the hyperbola $45^\circ$ so that its equation is $$x y = \frac12a^2 \tag{1}$$ whose asymptotes are simply the $x$- and $y$-axes. For $P=(p,q)$ on the curve, the midpoints to those axes are, respectively, $P_x(p/2,q)$ and $P_y(p,q/2)$, which satisfy $$x y = \frac12 p q = \frac14a^2 \tag{2}$$ Rotating back, we get the "standard form" equation to be $$x^2 - y^2 = \frac12 a^2 \tag{3}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .