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Consider an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$

We want to find the locus of the midpoints of all those chords whose length is constant $(=2c)$

Here's my approach:

Let the midpoint of the chord be $(h,k)$ So, the equation of the chord passing through it is given by:

$$T=S_1$$ $$\implies \frac{xh}{a^2} + \frac {yk}{b^2} -1 = \frac{h^2}{a^2} + \frac{k^2}{b^2}-1$$

Then, from this equation, I found x in terms of y, substituted it into the equation of ellipse.

Also, I found y in terms of x and substited it into the equation of ellipse.

Each of the substitution resulted in two quadratic equations in terms of x and y.

We can find sum and products of its roots to get these:

$$ x_1 + x_2 = \frac{2a^2 h k^2 + 2b^2 h^3}{a^2k^2 + b^2h^2}$$ $$ x_1 x_2 = \frac{(b^2h^2+a^2k^2)^2-a^4b^2k^2}{a^2b^2k^2 + b^4h^2}$$

$$ y_1 + y_2 = \frac{2b^2 h^2 k + 2a^2 k^3}{a^2k^2 + b^2h^2}$$ $$ y_1 y_2 = \frac{(b^2h^2+a^2k^2)^2-a^2b^4h^2}{a^2b^2h^2 + a^4k^2}$$

Here, $(x_1,y_1)$ and $(x_2,y_2)$ are actually the points where the the chord intersects the ellipse.

So, using distance formula,

$$2c = \sqrt{(x_1 - x_2)^2 + (y_1-y_2)^2}$$

Squaring both sides, and simplifying:

$$ 4c^2 = (x_1 + x_2)^2 + (y_1 + y_2)^2 - 4(x_1 x_2 + y_1 y_2)$$

Now, We can substitute the values of $x_1 + x_2, x_1x_2, y_1 +y_2,y_1y_2$ that we got earlier to finally get the answer.

However, I don't think that would be a "humane" thing to do.

Also, Please tell me if there are other methods to approach this problem.

Ayush Kumar
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2 Answers2

1

We can ease the work by not over simplifying terms such as

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}$$

Using results from my older post, we have

  • Pole and polar

$$\frac{x'x}{a^2}+\frac{y'y}{b^2}=1$$

  • Mid-point of polar chord

$$\begin{pmatrix} X \\ Y \end{pmatrix}= \frac{1}{\dfrac{x'^2}{a^2}+\dfrac{y'^2}{b^2}} \begin{pmatrix} x' \\ y' \end{pmatrix}$$

  • Chord length

$$\ell= \frac{2ab}{\dfrac{x'^2}{a^2}+\dfrac{y'^2}{b^2}} \sqrt{ \left( \dfrac{x'^2}{a^4}+\dfrac{y'^2}{b^4} \right) \left( \dfrac{x'^2}{a^2}+\dfrac{y'^2}{b^2}-1 \right)}$$

  • Inversion properties between pole and chord-centre

\begin{align} \begin{pmatrix} x' \\ y' \end{pmatrix} &= \frac{1}{\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}} \begin{pmatrix} X \\ Y \end{pmatrix} \\[5pt] \frac{x'^2}{a^2}+\frac{y'^2}{b^2} &= \frac{1}{\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}} \end{align}

  • The required locus with fixed chord length $\ell$

\begin{align} \ell &= \frac{2ab}{\dfrac{x'^2}{a^2}+\dfrac{y'^2}{b^2}} \sqrt{ \left( \frac{x'^2}{a^4}+\frac{y'^2}{b^4} \right) \left( \frac{x'^2}{a^2}+\frac{y'^2}{b^2}-1 \right) } \\[5pt] \ell &= \frac{2ab\sqrt{ \left( \dfrac{X^2}{a^4}+\dfrac{Y^2}{b^4} \right) \left( 1-\dfrac{X^2}{a^2}-\dfrac{Y^2}{b^2} \right)}}{ \sqrt{\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}}} \end{align}

Ng Chung Tak
  • 18,990
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Take the midpoint as $P(x_1,y_1)$ Chord through P is $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$$ The general distance form of this line is $$R(x,y)=(x_1+r\cos \theta,y_1+r\sin \theta)\\ \ \\ where \ \ \theta=-\arctan \frac{b^2x_1}{a^2y_1}$$

We need a point P such that at $r=c$ the point lies on the ellipse $$ \frac{(x_1+c\cos \theta)^2}{a^2}+\frac{(y_1+c\sin \theta)^2}{b^2}=1 \\ \ \\ (\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1)+2c(x_1\cos\theta+y_1\sin\theta)+\frac{c^2}{1+\tan^2\theta}(\frac{1}{a^2}+\frac{\tan^2\theta}{b^2})=0 \\ \ \\ (\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1)+\frac{2cx_1y_1(a^2-b^2)}{\sqrt{b^4x^2+a^4y^2}}+\frac{a^2b^2c^2}{b^4x^2+a^4y^2}(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2})=0 $$

I think you can clean it up a bit with eccentricity but this should be the locus

RandomGuy
  • 1,349