Solution:
Solution:
The points \(P(ap^2, 2ap)\) and \(Q(aq^2, 2aq)\), where \(p>0\) and \(q<0\), lie on the curve \(C\) with equation $$y^2= 4ax\,, $$ where \(a>0\,\). Show that the equation of the tangent to \(C\) at \(P\) is $$y= \frac 1 p \, x +ap\,.$$ The tangents to the curve at \(P\) and at \(Q \) meet at \(R\). These tangents meet the \(y\)-axis at \(S\) and \(T\) respectively, and \(O\) is the origin. Prove that the area of triangle \(OPQ\) is twice the area of triangle \(RST\).
A circle of radius \(a\) is centred at the origin \(O\). A rectangle \(PQRS\) lies in the minor sector \(OMN\) of this circle where \(M\) is \((a,0)\) and \(N\) is \((a \cos \beta, a \sin \beta)\), and \(\beta\) is a constant with \(0 < \beta < \frac{\pi}{2}\,\). Vertex \(P\) lies on the positive \(x\)-axis at \((x,0)\); vertex \(Q\) lies on \(ON\); vertex \(R\) lies on the arc of the circle between \(M\) and \(N\); and vertex \(S\) lies on the positive \(x\)-axis at \((s,0)\). Show that the area \(A\) of the rectangle can be written in the form \[ A= x(s-x)\tan\beta \,. \] Obtain an expression for \(s\) in terms of \(a\), \(x\) and \(\beta\), and use it to show that \[ \frac{\d A}{\d x} = (s-2x) \tan \beta - \frac {x^2} s \tan^3\beta \,. \] Deduce that the greatest possible area of rectangle \(PQRS\) occurs when \(s= x(1+\sec\beta)\) and show that this greatest area is \(\tfrac12 a^2 \tan \frac12 \beta\,\). Show also that this greatest area occurs when \(\angle ROS = \frac12\beta\,\).
In this question, you may assume that, if a continuous function takes both positive and negative values in an interval, then it takes the value \(0\) at some point in that interval.
The triangle \(ABC\) has side lengths \(\left| BC \right| = a\), \(\left| CA \right| = b\) and \(\left| AB \right| = c\). Equilateral triangles \(BXC\), \; \(CY\!A\) \hspace{0.0mm} and \(AZB\) are erected on the sides of the triangle \(ABC\), with~\(X\) on the other side of \(BC\) from \(A\), and similarly for \(Y\) and \(Z\). Points \(L\), \(M\) and \(N\) are the centres of rotational symmetry of triangles \(BXC\), \(CY\!A\) and \(AZB\) respectively.
Two sequences are defined by \(a_1 = 1\) and \(b_1 = 2\) and, for \(n \ge 1\), \begin{equation*} \begin{split} a_{n+1} & = a_n+ 2b_n \,, \\ b_{n+1} & = 2a_n + 5b_n \,. \end{split} \end{equation*} Prove by induction that, for all \(n \ge 1\), \[ a_n^2+2a_nb_n - b_n^2 = 1 \,. \tag{\(*\)}\]
A particle is projected at speed \(u\) from a point \(O\) on a horizontal plane. It passes through a fixed point \(P\) which is at a horizontal distance \(d\) from \(O\) and at a height \(d \tan \beta\) above the plane, where \(d>0\) and \(\beta \) is an acute angle. The angle of projection \(\alpha\) is chosen so that \(u\) is as small as possible.
Particles \(P_1\), \(P_2\), \(\ldots\) are at rest on the \(x\)-axis, and the \(x\)-coordinate of \(P_n\) is \(n\). The mass of \(P_n\) is \(\lambda^nm\). Particle \(P\), of mass \(m\), is projected from the origin at speed \(u\) towards \(P_1\). A series of collisions takes place, and the coefficient of restitution at each collision is \(e\), where \(0 < e <1\). The speed of \(P_n\) immediately after its first collision is \(u_n\) and the speed of \(P_n\) immediately after its second collision is \(v_n\). No external forces act on the particles.
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