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{\bf Question}

Find the co-ordinates of the point of intersection of the tangents
at the points with parameters $t_1, t_2$ of the parabola $x =
kt^2$, $y = 2kt$.  Prove that the tangents intersect at right
angles if and only they intersect on the directrix.

\vspace{.25in}

{\bf Answer}

$\ds x = kt^2 \hspace{.2in} y = 2kt \hspace{.2in} \frac{dy}{dx} =
\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2k}{2kt} = \frac{1}{t}
$

So the equation of the tangent is \begin{eqnarray*} y - 2kt & = &
\frac{1}{t} (x - kt^2) \\ yt & = & x+kt^2 \end{eqnarray*}

So $yt_1 = x + kt^2$ and $yt_2 = xkt_2^2$ intersect where
$y(t_1-t_2) = k(t_1^2 - t_2^2)$

i.e. where $y = k(t1 +t2)$

So $x = kt_1(t_1+t_2) - kt_1^2 = kt_1t_2$

The directrix has equation $x = -k$

So the intersection lies on the directirx if and only if $t_1t_2 =
-1$ i.e. $\ds \frac{1}{t_1} \cdot \frac{1}{t_2} = -1$ i.e. if the
tangents intersect orthogonally

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