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Given is: ..a point P on the directrix of a parabola with foci $F$.

What I want to show is: the tangents of a parabola given by $y=kx^2$ through $P = (x_0, y_0)$ are orthogonal and the line between the points, where the tangents touch the parabola, contains the foci F.

What I have done is:

  1. I need a tangent through $P$ that touches the parabola in $B = (a, ka^2)$. So I have $m = \frac{\delta x}{\delta y}=\frac{ka^2-y_0}{a-x_0} \rightarrow ka^2-y_0 =m(a-x_0).$

  2. However, with the first derivate of the parabola I get: $y' = 2ka$ and therefore $m = 2ka \rightarrow a = \frac{m}{2k}$.

  3. From 1 and 2 I get: $k\cdot \frac{m^2}{4k^2} -y_0 = m\cdot (\frac{m}{2k} - x_0)$. And after some steps: $0 = m^2 - 4kmx_0 + 4ky_0$.

  4. So because $y_0 < kx_0^2$ (because $P$ is external to the parabola!) I should get two solutions for this quadratic equation, which means 2 possible values $m_1, m_2$.

  5. $m_1 = 2(kx_0 - \sqrt{x_0^2 k^2 -ky_0})$ and $m_2 = 2(kx_0 +\sqrt{x_0^2 k^2 -ky_0})$.

  6. Finally I have $m_1 \cdot m_2 = 4ky_0 = -1 \leftrightarrow y_0 = \frac{-1}{4k}$ or in words: $P$ must be on the directrix! So if the tangents are orthogonal, P must be on the directrix. This should be enough for the first step?

  7. I now have to show that the line, that contains the 'touching points' $B = (a, ka^2)$ and a second points $C = (c, kc^2 )$ also contains the foci $F$. How can I now show this?

Many thanks for any hints and solutions (or corrections) :)

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    $\begingroup$ I don't understand your formulation of the problem: Shouldn't it be : "the tangents of a parabola given by ... through an external point ... are orthogonal IF AND ONLY IFthe line between the points, where the tangents touch the parabola, contains the FOCUS F." ? $\endgroup$ – Jean Marie May 15 '16 at 17:39
  • $\begingroup$ @JeanMarie Thank you for the comment: I edited something. Given was only the "Let there be a point on a directrix and a parabola with foci F. "Then I have to show what I wrote above: Show that the tangents through P are orthogonal AND the points, where the tangents touch the parabola, contains the focus F.. In fact, P was already given as "on the directrix". $\endgroup$ – Vazrael May 15 '16 at 17:50
  • $\begingroup$ See the related recent issue Prove that the directrix is tangent to the circles that are drawn on a focal chord of a parabola as diameter and my answer to this question. $\endgroup$ – Jean Marie May 15 '16 at 22:55
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First express the $x$ and $y$ coordinates in terms of $k$ and $m$. You already derived $a=m/{2k}$ for point $B$. Replace $m$ with $-1/m$ for the perpendicular tangent to get $c={-1}/{2km}$ for point $C$. Note that the $x$ coordinates are opposite-signed so line $BC$ is sure to intersect the $y$ axis.

Let $(x_B,y_B)$ be the coordinates of $B$ found above, and label the coordinates of $C$ analogously. Define weighting variables $w_B, w_C$ such that:

$w_B+w_C=1$

$x_Bw_B+x_Cw_C=0$

Then (0,y_F) is collinear with $B$ and $C$ by selecting

$y_Bw_B+y_Cw_C\equiv y_F$

The resulting value of $y_F$ will match the focus.

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