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{\bf Question}

Let P be a point in a parabola with focus S.  Let the tangent at P
meet the directrix at N.  Let M be the foot of the perpendicular
from P to the directrix.  Show that the angle $PSN$ is a right
angle, and use this to deduce the parabolic mirror property

\vspace{.25in}

{\bf Answer}

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${}$

The equation of the tangent to $x=kt^2 \, \, y = 2kt$ is $\ds yt =
x+kt^2$

This meets the directrix  where $\ds x=-k$ so $\ds y =
\frac{k}{t}(t^2 -1)$

${}$

$S = (k,0)$ so the gradient of SP is $\ds \frac{2kt}{kt^2 - k} =
\frac{2t}{t^2 -1}$

${}$

The gradient of NS is $\ds \frac{\frac{k}{t}(t^2-1)}{-k-k} =
-\frac{t^2-1}{2t}$

The product of the gradients is -1 so $N\hat{S}P = 90^\circ$

Now $MP=PS$ by definition of a parabola so the triangles $MPS$ and
$SPN$ are congruent (RHS)

Hence the parabolic mirror property is in diagram.







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