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So I was doing this question question The locus of middle points of parallel chords of hyperbola :$xy=c^2$

The locus of mid-points of parallel chords of a conic is called a diameter or so I was told. I derived it as :

  • Let $y=mx+c $ be the parallel chords
  • Then for hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ $$\implies \frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}=1$$
  • $x_1+x_2=\frac{2a^2mc}{b^2-a^2m^2}$ where $x_1,x_2$ are roots of equation as well as abscissa of points of intersection
  • Let $(h,k)$ be mid-points of chords. Then $$h=\frac{a^2mc}{b^2-a^2m^2}=\frac{a^2m(k-mh)}{b^2-a^2m^2}$$
  • Therefore the locus of mid-points is $$k=\frac{hb^2}{a^2m}\implies y=\frac{xb^2}{a^2m}$$
  • For rectangular hyperbola, $a=b\implies y=\frac{x}{m}$
  • Therefore $my-x=0$

However the correct answer according to the book is [Option-A]: $y+mx=0$

Why does the formula for general hyperbola fail here?
How to get the correct solution?

EDIT: HERE'S a screenshot of the possible answer?

Aurelius
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  • Its just names, so they use 1/m instead of m. – trula May 13 '24 at 21:28
  • But that makes the question useless? Since the options are $$y=+mx$$ $$y=-mx$$ $$x=+my$$ $$x=-my$$ – Aurelius May 13 '24 at 22:08
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    Exploiting the multiple-choice nature ... The hyperbola is symmetric about $y=x$, so that the mdpts of chords parallel to that line (which has slope $m=1$) trace a line with slope $-1$. This test case tells us that the slope changes sign (but is inconclusive about reciprocation), restricting our options to (a) $y=-mx$ or (d) $y=-\frac1m x$. Of course, (d) would mean that the locus is always perpendicular to the chords, which is probably too much to expect; indeed, it shouldn't be too hard to convince yourself that "steep" chords correspond to "steep" diameters, indicating answer (a). – Blue May 13 '24 at 23:03
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    Another observation: Since the parameter $c$ doesn't figure into any of the possible answers, you could take it to be something especially convenient ... like, say, zero. The hyperbola $xy=0$ is nothing more than the coordinate axes themselves; analyzing how the slopes of chords and their respective diameters relate becomes pretty straightforward in this case. – Blue May 13 '24 at 23:06
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    As to what went wrong with your own efforts: Note that you were given the hyperbola $xy=c^2$, which has the coordinate axes as its asymptotes. You analyzed $\frac{x^2}{a^2}-\frac{y^2}{a^2}=1$, effectively rotating the given hyperbola by $45^\circ$. ... Put another way, you didn't use the equation for the "general hyperbola"; you used the one for the "general origin-centered hyperbola with transverse axis parallel to the $x$-axis". The given hyperbola does not fit this description, since its transverse axis is the line $y=x$. ... Try doing what you did, but using the equation $xy=c^2$. – Blue May 13 '24 at 23:15
  • @Blue If $xy=0$ are the coordinate axes. Then does it mean that equation of any pair of lines is a special case of hyperbola – Aurelius May 14 '24 at 11:01
  • From this Question "The funny thing is I've always thought of conic sections this way (after I learned about them). A parabola is an ellipse with a focal point at infinity; it is also a hyperbola with a focal point at infinity. To get from an ellipse to a hyperbola, the point wraps around at infinity." So every conic is a special case of hyperbolas – Aurelius May 14 '24 at 11:05
  • @Aurelius No , that's a bad wording. The equation of any conic is given by the second degree equation, in which the coefficients of each terms contributes to decide whether that equation represents straight lines , circle , ellipse etc. – An_Elephant May 14 '24 at 11:11
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    @Aurelius: "Then does it mean that equation of any pair of lines is a special case of hyperbola?" Almost. Any pair of crossed lines is a special case of a hyperbola; a pair of parallel/coincident lines is a special parabola; a point is a special ellipse. The term-of-art for special here is "degenerate". – Blue May 14 '24 at 12:39
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    @Aurelius: "So every conic is a special case of hyperbolas" Yes! (It happens to be more common to say that every conic is special case of an ellipse ... but, of course, the upshot of the observation is that it doesn't matter which way you think about it.) See my animation here. ... This unified notion of conics should serve you very well, so keep it in the front of your mind throughout your studies. It underlies what's called "projective geometry". Congratulations on seeing it! :) – Blue May 14 '24 at 12:50

1 Answers1

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Let the midpoints of system of parallel chords of slope '$m$' be $P(h,k)$.

Then , the equation of that chord is given by :

$$T_1=S_1$$

$$\implies\frac{ xk+yh}{2}-c^2= hk-c^2 $$

The slope of above line is $\frac{-k}{h}$ , which is equal to $'m'$ as given in question. Hence the locus of midpoint is : $$m=\frac{-k}{h}\implies mx+y=0$$

Notations:

For any point $P(x_1,y_1)$ and conic $$S:ax^2+by^2+2hxy+2gx+2fy+c=0$$ following notations are very useful in analytic geometry:

$T_1: $The expression of conic after the following replacement : $$x^2\to xx_1$$ $$y^2\to yy_1$$ $$xy\to \frac{xy_1+yx_1}{2}$$ $$x\to \frac{x+x_1}{2} , y\to \frac{y+y_1}{2}$$

$S_1:$ The expression of conic after the following replacement $(x,y)\to(x_1,y_1)$

It is a well known result that chord with a given midpoint $P$ of any conic ( circle, ellipse , hyperbola and parabola) is given by $T_1=S_1$.

An_Elephant
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