<p>The diagram shows a curved mirror such that any ray of light emanating from the source located at the origin $O$ of a Cartesian coordinate system is always reflected parallel to the $x$-axis.</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/converted/1758545698789-flljqbypk3.png" width="400" class="aligncenter" /></figure>
<p>The ray emanating from $O$ strikes the curved mirror at $P$ with coordinates $\left( x,y \right)$. Let $\alpha $ be the acute angle which $OP$ makes with the tangent to the curved mirror at $P$ and let $\beta $ be the angle which $OP$ makes with the $x$-axis.</p>
<p>Assume that the law of reflection of light holds, in other words, the reflected ray also makes an angle of $\alpha $ with the tangent to the curved mirror at $P$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Express $\beta $ in terms of $\alpha $.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Write down expressions for $\tan \,\alpha $ and $\tan \beta $ in terms of $x$, $y$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$, whichever is appropriate.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Using the results in **(i)** and **(ii)**, prove that</span></div><div class="list-item-paragraph" style="text-align: center;">$\left[ 1-{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}} \right]\left( \frac{y}{x} \right)=2\frac{\mathrm{d}y}{\mathrm{d}x}$. (1)</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Use the substitution ${p = \frac{\mathrm{d}y}{\mathrm{d}x}}$ to show that the differential equation (1) can be reduced to the differential equation</span></div><div class="list-item-paragraph" style="text-align: center;">$y\frac{\mathrm{d}y}{\mathrm{d}x}+x=\pm \sqrt{{{x}^{2}}+{{y}^{2}}}$. (2)</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Use the substitution ${z = x^2 + y^2}$ to find the general solution of the differential equation (2).</span></div><p style="text-align: right;">[5]</p></li></ol>

The diagram shows a curved mirror such that any ray of light emanating from the source located at the origin $O$ of a Cartesian coordinate system is always reflected parallel to the $x$-axis.

The ray emanating from $O$ strikes the curved mirror at $P$ with coordinates $\left( x,y \right)$. Let $\alpha $ be the acute angle which $OP$ makes with the tangent to the curved mirror at $P$ and let $\beta $ be the angle which $OP$ makes with the $x$-axis.

Assume that the law of reflection of light holds, in other words, the reflected ray also makes an angle of $\alpha $ with the tangent to the curved mirror at $P$.

Write down expressions for $\tan \,\alpha $ and $\tan \beta $ in terms of $x$, $y$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$, whichever is appropriate.

Using the results in **(i)** and **(ii)**, prove that

$\left[ 1-{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}} \right]\left( \frac{y}{x} \right)=2\frac{\mathrm{d}y}{\mathrm{d}x}$. (1)

Use the substitution ${p = \frac{\mathrm{d}y}{\mathrm{d}x}}$ to show that the differential equation (1) can be reduced to the differential equation

$y\frac{\mathrm{d}y}{\mathrm{d}x}+x=\pm \sqrt{{{x}^{2}}+{{y}^{2}}}$. (2)

Use the substitution ${z = x^2 + y^2}$ to find the general solution of the differential equation (2).

RI Differential Equations Tutorial Q10

RI Differential Equations Tutorial Q10

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