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Question: Prove that the directrix is tangent to the circles that are drawn on a focal chord of a parabola as diameter.

Here is a picture; enter image description here


What I have attempted;

Let the parabola be $y^2=4ax$

Hence the focus will be at $(a,0)$

Let the focal chord be $y = m(x-a) $

Subbing in $y^2=4ax$

$$y^2=4ax$$

$$ \Leftrightarrow (m(x-a))^2 = 4ax $$

$$ \Leftrightarrow m^2 (x^2-2ax+a^2) = 4ax $$

$$\Leftrightarrow m^2x^2 - 2am^2x + m^2a^2 - 4ax = 0 $$

$$ \Leftrightarrow m^2x^2 -(2am^2+4a)x + m^2a^2 = 0 $$

If $x_1$ and $x_2$ are roots then

$$ x_1 + x_2 = \frac{2am^2+4a}{m^2}$$

$$ \therefore x_1 + x_2 = 2a + \frac{4a}{m^2} $$

and $$ x_1 \cdot x_2 = \frac{m^2a^2}{m^2} $$

$$ \therefore x_1 \cdot x_2 = a^2 $$

Corresponding

$$y_1 + y_2 = m(x_1 - a + x_2 - a)$$

$$y_1 + y_2 = m(x_1 + x_2 - 2a)$$

$$y_1 + y_2 = m(2a + \frac{4a}{m^2} - 2a)$$

$$ \therefore y_1 + y_2 = \frac{4a}{m} $$

$$ y_1 \cdot y_2 = m^2(x_1-a)(x_2-a) $$

$$y_1 \cdot y_2 = m^2(x_1x_2 - a(x_1+x_2) + a^2) $$

$$ y_1 \cdot y_2 = m^2( a^2 - a^2(2 + \frac{4}{m^2}) + a^2) $$

$$ y_1 \cdot y_2 = m^2 (\frac{-4a^2}{m^2}) $$

$$ y_1 \cdot y_2 = -4a^2 $$

Now consider

$$ (x_1 - x_2)^2 = (x_1+x_2)^2 - 4x_1x_2 $$

$$ (x_1 - x_2)^2 = a^2(2 + \frac{4}{m^2})^2 - 4a^2 $$

$$ (x_1 - x_2)^2 = a^2(4+\frac{16}{m^2} + \frac{16}{m^4}) - 4a^2 $$

$$ (x_1 - x_2)^2= \frac{16a^2}{m^2} + \frac{16a^2}{m^4} $$

and

$$ (y_1 - y_2)^2 = (y_1+y_2)^2 - 4y_1y_2 $$

$$(y_1 - y_2)^2 = (\frac{4a}{m})^2 -4 \cdot -4a^2 $$

$$(y_1 - y_2)^2 = \frac{16a^2}{m^2} + 16a^2 $$

Therefore

$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = \frac{16a^2}{m^2} + \frac{16a^2}{m^4} + \frac{16a^2}{m^2} + 16a^2 $$

$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = 16a^2(\frac{1}{m^4} + \frac{2}{m^2} + 1) $$

$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = 16a^2(\frac{1}{m^2} + 1)^2 $$

Hence diameter of the circle is given as

$$ D = \sqrt{16a^2(\frac{1}{m^2} + 1)^2} $$

$$ \therefore D = 4a(\frac{1}{m^2} + 1) $$

Distance from centre of directrix is the $x$ coordinate $+a$

$$= a + \frac{2a}{m^2} + a $$

$$= 2a + \frac{2a}{m^2} $$

$$= 2a(1+\frac{1}{m^2}) $$

So distance is $2a(1+\frac{1}{m^2}) $

Also notice that the radius of the circle is given as $R = 2a(\frac{1}{m^2} + 1) $

Which equals the distance from centre to the directrix hence the directrix must be tangent to the circle.


Could someone please check my proof and tell me if I am correct or not (correct my working and tell me where i went wrong) or also provide me with an alternative way of approaching this question?

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Here is a simple alternative way, fully geometrical.

Have a look at the following picture, with $M_1,M_2$ on parabola with focus $F$ and directrix $D$, $H_1, H_2, H$ the orthogonal projections on D of $P_1, P_2, F$ resp.

Let $r_k:=M_kH_k=M_kF \ (k=1,2)$. Let $C$ be the midpoint of $M_1M_2$, i.e., the center of the circle with diameter $M_1M_2$. The radius of this circle is $r=\dfrac{r_1+r_2}{2}$.

In trapezoid $M_1,H_1,H_2,M_2$, consider line segment $[FH]$ joining midpoints $C$ and $H$ of line segments $M_1M_2$ and $H_1H_2$ resp. The length of $[FH]$ is the mean $\dfrac{r_1+r_2}{2}$ of the lengthes $r_1$ and $r_2$ of $[M_1H_1]$ and $[M_2H_2]$ resp., i.e., the radius of the circle, as desired.

enter image description here

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  • $\begingroup$ I have changed a little my proof. Say if it is convenient for you. A comment: in many cases, for conical sections, it is advantageous to see first if geometrical arguments can be used (sometimes it is not possible!), to save time compared to very tedious analytical geometry proofs, that, above all, do not bring a supplementary understanding.. $\endgroup$ – Jean Marie May 6 '16 at 16:04
  • $\begingroup$ Thank you for that , I was wondering if my proof is correct as well? $\endgroup$ – bigfocalchord May 6 '16 at 19:56
  • $\begingroup$ Yes, I have checked the different lines. $\endgroup$ – Jean Marie May 6 '16 at 21:24
  • $\begingroup$ @JeanMarie thank you, one last question is there a way you could share the geogebra file (i have had trouble drawing this) $\endgroup$ – bigfocalchord Jun 28 '16 at 23:13
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An alternate way (probably shorter) would be to take two points as $(at_1^2,2at_1), (at_2^2,2at_2)$ as the ends of the focal chord.

As this chord passes through focus, we obtain $t_1t_2=-1$ Now, the equation of circle can be written in diametric form as:

$$(x-at_1^2)\left(x-\frac{a}{t_1^2}\right)+(y-2at_1)\left(y+\frac{2a}{t_1}\right)=R^2$$ Where $R$ can be written using $t_1$ and $t_2$.

Now you can differentiate this at $x=-a$ and check for the slope of the tangent.

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  • $\begingroup$ Very witty! Something that might be non evident for the reader is the $t_1t_2=-1$. Maybe, one could say that the alignment of points $M_1,M_2$ and $F$ can be translated into the fact that $$det \ \begin{bmatrix}at_1^2&2at_1&1\\ at_2^2&2at_2&1\\ a&0&1\end{bmatrix}=0$$ $\endgroup$ – Jean Marie May 6 '16 at 16:11
  • $\begingroup$ @JeanMarie true you say the area of the triangle would be $0$, Although I did it by equating the slopes, $$\frac{2at_1}{at_1^2-a}=\frac{-2at_2}{a-at_2^2}$$. $\endgroup$ – Nikunj May 7 '16 at 7:30
  • $\begingroup$ For the diametric form why is the radius zero? $\endgroup$ – bigfocalchord Jun 28 '16 at 5:14
  • $\begingroup$ @dydxx idk what I was thinking at that time, it will not be $0$ $\endgroup$ – Nikunj Jun 28 '16 at 20:26
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    $\begingroup$ @dydxx yup, (it will be ugly) $\endgroup$ – Nikunj Jun 28 '16 at 20:47

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