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xXnukerrrXx · Sep 5, 2011

From 9I Year 11 Cambridge - Alternative proof to reflection property. Part d and e stuck. Please help

rolpsy · Sep 6, 2011

$P(x_{1}, \ y_{1}) \ \ S(0, \ a) \ \ Q \left ( \frac{2ay_{1}}{x_{1}} , \ 0 \right ) \ \ R(0 , \ -y_{1})$
d)
$\begin{align*} m_{SQ}&=\frac{0-a}{\frac{2ay_{1}}{x_{1}}-0}\\ &=\frac{-a\times x_{1}}{2ay_{1}}\\ &=-\frac{x_{1}}{2y_{1}}\\\\ m_{PQ}&=\frac{y_{1}-0}{x_{1}-\frac{2ay_{1}}{x_{1}}}\\ &=\frac{y_{1}\times x_{1}}{(x_{1})^{2}-2ay_{1}}\\ &=\frac{x_{1}y_{1}}{x_{1}-\frac{(x_{1})^2}{2}} \ \ \textup{ .. as } (x_{1})^{2} = 4ay_{1}, \ \ \textup{ so } \ \ 2ay_{1}=\frac{(x_{1})^2}{2}\\ &=\frac{x_{1}y_{1}}{\frac{(x_{1})^2}{2}}\\ &=\frac{2y_{1}}{x_{1}}\\ &\therefore m_{SQ} \times m_{PQ} = -1\\ &\therefore SQ \perp PQ \end{align*}$

$\begin{align*} PQ^{2}&=\left ( x_{1}-\frac{2ay_{1}}{x_{1}} \right )^{2}+(y_{1}-0)^{2}\\ &=(x_{1})^{2}-4ay_{1}+\left (\frac{2ay_{1}}{x_{1}} \right )^{2} + (y_{1})^{2}\\ &=\left (\frac{2ay_{1}}{x_{1}} \right )^{2} + (y_{1})^{2} \ \ \textup{ .. as } (x_{1})^{2} = 4ay_{1} \\\\ RQ^2&=\left (\frac{2ay_{1}}{x_{1}} -0\right )^{2}+(0-(-y_{1}))^2\\ &=\left (\frac{2ay_{1}}{x_{1}} \right )^{2} + (y_{1})^{2}\\ &=PQ^2\\ &\therefore RQ=PQ\\ &\therefore \textup{The tangent at the vertex (the x-axis) bisects PR} \end{align*}$

edit:
e)

$\begin{align*} \textup{In } &\triangle SQR \textup{ and } \triangle SQP,\\ &\textup{S } \ \ SQ \textup{ is common}\\ &\textup{A } \ \ \angle SQR=\angle SQP=90^{\circ} \ \ \textup{ .. as } SQ \perp PQ \\&\textup{S } \ \ RQ=PQ \ \ \textup{ .. as the tangent at the vertex bisects PR}\\ \therefore \ \ &\triangle SQR \cong \triangle SQP \ \ (SAS) \end{align*}$

i'm not sure what e) is asking :S
i think it may be ∠SRQ = ∠SPQ (corresponding angles in congruent triangles △SQR ≅ △SQP equal)

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xXnukerrrXx

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