RI Differential Equations Tutorial Q10

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 \(P\).

1. Express \(\beta \) in terms of \(\alpha \).[2]
2. 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.[2]
3. 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)[2]
4. 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)[2]
5. Use the substitution \({z = x^2 + y^2}\) to find the general solution of the differential equation (2).[5]