Problems

Solution:

1. The line \(AB\) has equation: \begin{align*} && \frac{y+f(b)}{x-b} &= \frac{f(a)+f(b)}{a-b} \\ \Rightarrow && \frac{f(b)}{m-b} &= \frac{f(a)+f(b)}{a-b} \\ \Rightarrow && m &= \frac{a-b}{f(a)+f(b)}f(b) + b \\ &&&= \frac{af(b)+bf(a)}{f(a)+f(b)} \end{align*}
2. Suppose \(f(x) = \sqrt{x}\) then \begin{align*} m &= \frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}} \\ &= \frac{\sqrt{ab}(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}} \\ &= \sqrt{ab} \end{align*} Suppose \(f(x) = x^{-n}\) then \begin{align*} m &= \frac{a b^{-n}+ba^{-n}}{a^{-n}+b^{-n}} \\ &= \frac{a^{n+1}+b^{n+1}}{b^n + a^n} \\ \end{align*}
3. Without loss of generality, we can scale \(g_1(x)\) and \(g_2(x)\) so that \(g_1(a) = g_2(a)\) and \(m\) won't change for either of them. Then since \(\frac{g_1(b)}{g_2(b)} < 1\) (this function is decreasing) our line connecting \((a,g_i(a))\) and \((b,-g_i(b))\) must interect the axis first for \(g_2\), in particular \(M_1 > M_2\). Suppose \(g_1(x) =1, g_2(x) = \sqrt{x}, g_3(x) = x^{-1}\), the notice that \(\frac{g_1(x)}{g_2(x)} =\frac{g_2(x)}{g_3(x)}= x^{-1/2}\) are decreasing, therefore: \begin{align*} \frac{a+b}{1+1} &> \sqrt{ab} > \frac{1+1}{a^{-1}+b^{-1}} \\ \frac{a+b}{2} &> \sqrt{ab} > \frac{2ab}{a+b} \\ \end{align*}
4. We must have: \begin{align*} && p(c-a)^3 &= f(a) \\ && p(c-b)^3 &= -f(b) \\ \Rightarrow &&\left ( \frac{c-a}{c-b} \right)^3 &= -\frac{f(a)}{f(b)} \\ \Rightarrow && \frac{c-a}{b-c} &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \\ \Rightarrow && c-a &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}(b-c)\\ \Rightarrow && c \left (1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \right) &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a \\ \Rightarrow && c &= \frac{\left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a}{1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}} \\ &&&= \frac{b[f(a)]^\tfrac13+a[f(b)]^\tfrac13}{[f(a)]^\tfrac13+[f(b)]^\tfrac13} \end{align*} We have that \(\frac{c-a}{b-c} = \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \) and \(\frac{m-a}{b-c} = \frac{f(a)}{f(b)}\). Since \(f\) is decreasing, \(\frac{f(a)}{f(b)} > 1\) and so \(\left (\frac{f(a)}{f(b)} \right)^{\tfrac13} < \frac{f(a)}{f(b)}\), therefore \(m > c\).

Solution:

1. Let \(k = 1\), then \begin{align*} \dot{x} &= - x - y \\ \dot{y} &= x - y \\ \dot{x}-\dot{y} &= -2x \\ \ddot{x} &= -\dot{x}-\dot{y} \\ &= -\dot{x} - (\dot{x}+2x) \\ &= -2\dot{x}- 2x \\ \dot{x}+\dot{y} &= -2y \\ \ddot{y} &= \dot{x}-\dot{y} \\ &= -2y-2\dot{y} \end{align*} So we have an auxiliary equation for \(x\) and \(y\) which is \(\lambda^2 + 2 \lambda+2 = 0 \Rightarrow \lambda = -1 \pm i\). Therefore \(x = Ae^{-t} \cos t + B e^{-t} \sin t, y = Ce^{-t}\cos t + De^{-t} \sin t\). We also must have that, \(A = 1, C = 0\), so \(x = e^{-t} \cos t + Be^{-t} \sin t\) and \(y = De^{-t} \sin x\). \begin{align*} \dot{y} &= -De^{-t} \sin t +De^{-t} \cos t \\ &= e^{-t} \cos x + Be^{-t} \sin t- De^{-t} \sin t \\ \end{align*} therefore \(B = 0, D = 1\) and \(x = e^{-t} \cos t, y = e^{-t} \sin t\)
   

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   \begin{align*} y &= e^{-t} \sin t \\ \dot{y} &= -e^{-t} \sin t + e^{-t} \cos t \\ \dot{x} &= e^{-t} \cos t -e^{-t} \sin t \end{align*}
   

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2. \begin{align*} \dot{x} = -x \\ \dot{y} = x-y \end{align*} So \(x = e^{-t}\). \(\dot{y} + y = e^{-t}\) so \(y = (t+B)e^{-t}\) and so \(y =te^{-t}\).
   

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Solution: Note that \begin{align*} && \frac{\d a \sec \theta}{\d \theta} &= a \sec \theta \tan \theta \\ && \frac{\d b \tan \theta}{\d \theta} &= b \sec^2 \theta \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{b \sec^2 \theta}{a \sec \theta \tan \theta} \\ &&&= \frac{b}{a} \frac{1}{\sin \theta} \\ \Rightarrow && \frac{y-b \tan \theta}{x - a \sec \theta} &= \frac{b}{a} \frac{1}{\sin \theta} \\ \Rightarrow && a \sin \theta y - ab \tan \theta \sin \theta &= bx -ab \sec \theta \\ \Rightarrow && bx-ay\sin \theta &= ab \sec x (1 - \sin ^2 \theta) \\ &&&= ab \cos \theta \end{align*}

1. \begin{align*} S: &&& \begin{cases} bx-ay &= 0 \\ bx-ay \sin \theta &= ab \cos \theta \end{cases} \\ \Rightarrow && ay(1-\sin \theta) &= ab \cos \theta \\ \Rightarrow && y &= \frac{b \cos \theta}{1-\sin \theta} \\ &&x &=\frac{a\cos \theta}{1-\sin \theta} \\ T: &&& \begin{cases} bx+ay &= 0 \\ bx-ay \sin \theta &= ab \cos \theta \end{cases} \\ \Rightarrow && ay(1+\sin \theta) &= -ab \cos \theta \\ \Rightarrow && y &= \frac{-b \cos \theta}{1+\sin \theta} \\ &&x &=\frac{a\cos \theta}{1+\sin \theta} \\ M: && x &= \frac{a \cos \theta}{2} \frac{2}{1-\sin^2 \theta} \\ &&&= a \sec \theta \\ && y &= \frac{b \cos \theta}{2} \frac{2 \sin \theta}{1-\sin^2 \theta} \\ &&&= b \tan \theta \end{align*} The midpoint of \(ST\) is the same as \(P\).
2. The tangents are perpendicular, therefore \(\frac{b}{a} \cosec \theta = - \frac{a}{b} \sin \phi\), ie \(b^2 = -a^2 \sin \phi \sin \theta\) The will intersect at: \begin{align*} &&& \begin{cases} bx - ay \sin \theta &= ab \cos \theta \\ bx - ay \sin \phi &= ab \cos \phi \end{cases} \\ \Rightarrow && ay ( \sin \theta - \sin \phi) &= ab(\cos \phi - \cos \theta) \\ \Rightarrow && y &= \frac{b(\cos \phi - \cos \theta)}{(\sin \theta - \sin \phi)} \\ && y^2 &= \frac{-a^2 \sin \phi \sin \theta (\cos\phi - \cos \theta)^2}{(\sin \theta - \sin \phi)^2} \\ \Rightarrow && bx(\sin \phi - \sin \theta) &= ab(\cos \theta \sin \phi - \cos \phi \sin \theta) \\ \Rightarrow && x &= \frac{a(\cos \theta \sin \phi - \cos \phi \sin \theta)}{\sin \phi - \sin \theta} \\ &&&= \frac{a^2(\cos \theta \sin \phi - \cos \phi \sin \theta)^2}{(\sin \phi - \sin \theta)^2} \end{align*} Therefore \begin{align*} && x^2+y^2 &= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\cos \theta \sin \phi- \cos \phi \sin \theta)^2 - \sin \phi \sin \theta (\cos\phi - \cos \theta)^2 \r \\ &&&= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\sin \phi - \sin \theta)(\cos^2 \theta \sin \phi - \sin \theta \cos^2 \phi) \r \\ &&&=a^2-b^2 \end{align*}

Solution: \begin{align*} && 2yy' &= 4a \\ \Rightarrow && y' &= \frac{2a}{y} = \frac{2a}{2ap} = \frac1p \\ \Rightarrow && \frac{y-2ap}{x-ap^2} &= \frac1p \\ \Rightarrow && y &= \frac1p x +ap \end{align*} The other tangent will be \(y = \frac1qx+aq\) \begin{align*} &&& \begin{cases} py-x &= ap^2 \\ qy - x &= aq^2 \end{cases} \\ \Rightarrow && y(p-q) &= a(p^2-q^2) \\ \Rightarrow && y &= a(p+q) \\ && x &= apq \end{align*} Therefore \(R(apq, a(p+q)), S(0, ap), T(0, aq)\).

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Solution:

1. \begin{align*} && 2y \frac{\d y}{\d x} &= 4a \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{2a}{y} \end{align*} Therefore we must have \begin{align*} && \underbrace{-\frac{2aq}{2a}}_{\text{gradient of normal}} &= \underbrace{\frac{2ap-2aq}{ap^2-aq^2}}_{\Delta y / \Delta x} \\ \Rightarrow && -q &= \frac{2}{p+q} \\ && 0 &= 2 + pq+q^2 \end{align*}
2. We must have that \(q,r\) are the two roots of \(x^2+px+2 = 0\) \(QR\) has the equation: \begin{align*} && \frac{y-2aq}{x-aq^2} &= \frac{2ar-2aq}{ar^2-aq^2} \\ \Rightarrow && \frac{y-2aq}{x-aq^2} &= \frac{2}{r+q} \\ \Rightarrow && y &= \frac{2}{q+r}(x-aq^2) +2aq \\ && y &= -\frac{2}{p}x+2a\left(q-\frac{q^2}{q+r} \right) \\ &&y&= -\frac{2}{p}x+2a \frac{qr}{q+r} \\ && y &= -\frac{2}{p}x - 2a \frac{2}{p} \\ && y & = -\frac{2}{p}(x+2a) \end{align*} Therefore the point \((-2a,0)\) lies on all such lines.
3. \(OP\) has equation \(y = \frac{2}{p} x\) \begin{align*} && y &= \frac{2}{p} x \\ && y & = -\frac{2}{p}(x+2a) \\ && 2y &= -\frac{4a}{p} \\ \Rightarrow && y &= -\frac{2a}{p} \\ && x &= -a \end{align*} Therefore \(T\left (-a, -\frac{2a}{p} \right)\) always lies on the line \(x = -a\) The distance to the \(x\)-axis from \(T\) is \(\frac{2a}{|p|}\). We need to show that \(p\) can't be too small. Specifically \(x^2+px+2 = 0\) must have \(2\) real roots, ie \(\Delta = p^2-8 \geq 0 \Rightarrow |p| \geq 2\sqrt{2}\), ie \(\frac{2a}{|p|} \leq \frac{2a}{2\sqrt{2}} = \frac{a}{\sqrt{2}}\) as required.

Solution: The bullet fired at time \(t\) will hit the ground at time \(t+\frac{2u \sin (\frac13\pi - \lambda t)}{g}\). To find the last time a bullet hits the ground, we can differentiate, noting that \begin{align*} && T(t) &= t + \frac{2u \sin \alpha}{g} \\ \Rightarrow && T'(t) &= 1 - \frac{2u\lambda}{g} \cos \alpha \\ && T''(t) &= \frac{2u \lambda^2}{g} \sin \alpha > 0 \end{align*} If \(k = \frac{g}{2\lambda u} \in [\frac12, \frac12\sqrt{3}]\) then notice that this turning point is always achieved, and will be a maximum. It will be when \(\cos \alpha = k, \sin \alpha = \sqrt{1-k^2}\). The distance will be \(u \cos \alpha \cdot \frac{2 u \sin \alpha}{g} = \frac{2ku^2\sqrt{1-k^2}}{g}\). If \(k < \frac12\) then the last bullet to hit the ground will be the last bullet fired, ie \(\frac{2u^2 \sin \frac16\pi \cos \frac16\pi}{g} = \frac{u^2 \sin \frac13 \pi}{g} = \frac{\sqrt{3}u^2}{2g}\)

Solution:

1. \(L_a : \frac{y}{x-a} = \frac{1-a-0}{0-a} = \frac{a-1}{a} \Rightarrow ay+(1-a)x = a(1-a)\) \begin{align*} && ay + (1-a)x &= a(1-a) \\ && by + (1-b)x &= b(1-b) \\ \Rightarrow && aby + b(1-a)x &= ba(1-a) \\ && aby + a(1-b)x &= ab(1-b) \\ \Rightarrow && (b-a)x &= ab(b-a) \\ \Rightarrow && x &= ab \\ && y &= \frac{a-1}{a} \cdot a(b-1) \\ &&&= (1-a)(1-b) \end{align*}
2. As \(a \to b\), \(x \to a^2, y \to 1-2a+a^2 =(1-a)^2 = (1-\sqrt{x})^2\)
3. \(\frac{\d y}{\d x} = 2(1-\sqrt{x})\cdot \left (-\tfrac12 \frac{1}{\sqrt{x}} \right) = 1 - \frac{1}{\sqrt{x}}\). Therefore the tangent when \(x = c^2, y = (1-c)^2\) is \begin{align*} && \frac{y-(1-c)^2}{x-c^2} &= 1 - \frac{1}{c} \\ \Rightarrow && cy + (1-c)x &= c(c-1)+c(1-c)^2 \\ &&&= c(1-c) \end{align*} Which is an equation of the form \(L_c\)

Solution: \begin{align*} && v_{\uparrow} &= U\sin \theta - g t \\ \Rightarrow && T_H &= \frac{U \sin \theta}{g} \\ \\ && s_{\uparrow} &= U \sin \theta t - \frac12 g t^2 \\ \Rightarrow && 0 &= U\sin \theta T_L - \frac12 g T_L^2 \\ && T_L &= \frac{2 U \sin \theta}{g} \end{align*} \(T = U\cos \theta / (kg)\) is the point when the particle's horizontal motion is reversed.

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Solution:

1. Suppose we have the shortest distance between the two curves, and the path between the points is not a normal to both curves. Then we could shift the endpoints to reduce the distance. (Assuming we're not at a point of intersection). Therefore, the normal to the curves must be the same (or in other words) the gradients of the curves must be the same. ie we are at a point where \(2y y' = 4a\) we must have \(y' = m\), so \(y = \frac{2a}{m}\) and \(x = \frac{a}{m^2}\) and the distance from this point to the line \(y=mx+c\) is \(\frac{|m \frac{a}{m^2} - \frac{2a}{m}+c|}{\sqrt{m^2+1}} = \frac{|mc-a|}{m\sqrt{m^2+1}} = \frac{mc-a}{m\sqrt{m^2+1}}\). If \(mc \leq a\) then we find \(\frac{a-mc}{m\sqrt{m^2+1}}\) However, we must check that the two curves do not intersect (otherwise the closest distace is \(0\)). ie we need to check if \((mx+c)^2 = 4ax\) has any solutions, this quadratic has discriminant \((2mc-4a)^2 - 4 \cdot m^2 \cdot c^2 = 16a^2-16amc = 16a(a-mc)\) which is clearly greater than \(0\) when \(a \geq mc\). Therefore the shortest distance in this case is \(0\).
2. The distance squared between the point \((p,0)\) and a point of the form \((at^2,2at)\) is \(D^2 = (at^2-p)^2+4a^2t^2 = a^2t^4+(4a^2-2ap)t^2+p^2\) \begin{align*} && \frac{D^2}{a^2} &= t^4 + 2\left(2-\frac{p}{a}\right)t^2 + \frac{p^2}{a^2} \\ &&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 + \frac{p^2}{a^2} - \left (2-\frac{p}{a} \right)^2 \\ &&&= \left (t^2 - \left (\frac{p}{a}-2 \right)\right)^2 +\frac{4p}{a} -4 \\ \end{align*} Therefore if \(2 \leq \frac{p}{a}\) then we can find a \(t\) such that we attain the minimum for \(D^2/a^2\) of \(\frac{4p}{a}-4\) and so \(D = \sqrt{4pa-4a^2} = 2\sqrt{a(p-a)}\) . If not the smallest value will be when \(t = 0\) and we will have \(|p|\) Now consider all the lines joining points on the parabola to the centre of the circle. The shortest distance from the parabola to the circle will be normal to the circle and therefore will also be a line through the center. Therefore we need only consider the shortest distance from \((p,0)\) to the parabola \(-b\). Case 1: If \(p \geq 2a\) we have \(2\sqrt{a(p-a)} - b\) or \(0\) if \(b \geq 2\sqrt{a(p-a)}\) Case 2: If \(p < 2a\) we have \(p-b\) or \(0\) if \(b \geq p\)

Solution: \begin{align*} && y &= bx-ba \\ && 1 &= x^2 + y^2 \\ \Rightarrow && 1 &= x^2 + b^2(x-a)^2 \\ \Rightarrow && 0 &= (1+b^2)x^2-2ab^2x+b^2a^2-1 \end{align*} This will have roots which sum to \(\frac{2ab^2}{1+b^2}\), therefore \(M = \left ( \frac{ab^2}{1+b^2}, \frac{ab^3}{1+b^2}-ba \right)=\left ( \frac{ab^2}{1+b^2}, \frac{-ba}{1+b^2} \right)\)

1. Since \(b\) is fixed so is \(\frac{b}{1+b^2} = t\) and all the points are \((bta, -ta)\), ie \(x = -by\). If \(a \in [-1,1]\) we are ranging on the points \((bt, -t)\) to \((-bt, t)\) which is a distance of \begin{align*} && d &= \sqrt{(bt+bt)^2+(-2t)^2} \\ &&&= \sqrt{4(b^2+1)t^2} \\ &&&=2 \sqrt{(b^2+1)\frac{b^2}{(b^2+1)^2}} \\ &&&= \frac{2b}{\sqrt{b^2+1}} \end{align*}
   

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2. • If \(a\) is fixed we have \(\left ( \frac{ab^2}{1+b^2}, -\frac{ba}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= - b \\ \Rightarrow && y &= \frac{a\frac{x}{y}}{1 + \frac{x^2}{y^2}} \\ \Rightarrow && y^2 \left ( 1 + \frac{x^2}{y^2} \right) &= ax \\ \Rightarrow && x^2-ax + y^2 &= 0 \\ \Rightarrow && \left (x - \frac{a}{2} \right)^2 + y^2 &= \frac{a^2}{4} \end{align*} Therefore we will end up with a circle centre \((\tfrac{a}{2}, 0)\) going through the origin.
     

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   • If \(ab = 1\), we have \(\left ( \frac{b}{1+b^2}, -\frac{1}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= -b \\ \Rightarrow && y &= -\frac{1}{1+\frac{x^2}{y^2}} \\ \Rightarrow && y + \frac{x^2}{y} &= - 1 \\ \Rightarrow && y^2 +y+ x^2 &= 0 \\ \Rightarrow && \left ( y + \frac12 \right)^2 + x^2 &= \frac14 \end{align*}
     

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Solution: \begin{align*} && \frac{\d y}{\d x} &= -\frac1{2x^2} \\ \Rightarrow && \frac{y - \frac{1}{2p}}{x - p} &= - \frac{1}{2p^2} \\ \Rightarrow && y - \frac1{2p} &= -\frac{1}{2p^2}x +\frac1{2p} \\ \Rightarrow && y +\frac{1}{2p^2}x &= \frac1p \\ \Rightarrow && 2p^2 y + x &= 2p\\ \Rightarrow && 2q^2 y + x &= 2q \\ \Rightarrow && (p^2-q^2)y &= p-q \\ \Rightarrow && y &= \frac{1}{p+q} \\ && x &= \frac{2pq}{p+q} \end{align*} \begin{align*} \text{normal}: && \frac{y-\frac1{2p}}{x-p} &= 2p^2 \\ \Rightarrow && y - \frac1{2p} &= 2p^2x - 2p^3 \\ \Rightarrow && 2py -4p^3x &= 1-4p^4 \\ \Rightarrow && 2qy -4q^3x &= 1-4q^4 \\ pq = \tfrac12: && y - 2p^2 x &= q-2p^3 \\ && y - 2q^2 x &= p-2q^3 \\ \Rightarrow && (2q^2-2p^2)x &= q-p +2q^3-2p^3 \\ &&&= (q-p)(q+p+2q^2+1+2p^2) \\ \Rightarrow && x &= \frac{1+2(p^2+q^2)+1}{2(p+q)} \\ &&&= \frac{1+2(p^2+q^2+2pq-1)+1}{2(p+q)} \\ &&&= p+q\\ && y &= 2p^2 \left ( p+q \right) + q - 2p^3 \\ &&&= p+q \end{align*} So \(N(p+q, p+q)\) and \(T\left (\frac{1}{p+q}, \frac{1}{p+q} \right)\), so both points lie on \(y = x\). \[ OT \cdot ON = \frac{\sqrt{2}}{p+q} \cdot (p+q)\sqrt{2} = 2 \] which is clearly constant.

Solution: The tangent at \((at^2, 2at)\) can be found \begin{align*} && \frac{\d y}{\d x} &= \frac{\dot{y}}{\dot{x}} \\ &&&= \frac{2a}{2at} = \frac1t \\ \Rightarrow && \frac{y-2at}{x-at^2} &= \frac1t \\ \Rightarrow && ty -x &= at^2 \\ \\ PT: && py - x &= ap^2 \\ QT: && qy - x &= aq^2 \\ \Rightarrow && (p-q)y &= a(p^2-q^2) \\ \Rightarrow && y &= a(p+q) \\ && x &= aq(p+q) - aq^2 \\ &&&= apq \end{align*} By the cosine rule: \begin{align*} && TP^2 &= FT^2 + FP^2 - 2 \cdot FT \cdot FP \cdot \cos \phi \\ && (apq - ap^2)^2 + (a(p+q)-2ap)^2 &= (a-apq)^2+(a(p+q))^2 + \\ &&&\quad + (a-ap^2) + (2ap)^2 - 2 \cdot FT \cdot FP \cdot \cos \phi \\ \Rightarrow && a^2p^2(q-p)^2 + a^2(q-p)^2 &= a^2(1-pq)^2+a^2(p+q)^2 + \\ &&&\quad + a^2(1-p^2)^2+4a^2p^2 - 2 \cdot FT \cdot FP \cdot \cos \phi \\ && a^2(p^2+1)(q-p)^2 &= a^2(1+p^2)(1+q^2) + a^2(1+p^2)^2 + \\ &&&\quad - 2 \cdot a^2(1+p^2)\sqrt{(1+p^2)(1+q^2)} \cos \phi \\ \Rightarrow && \cos \phi &= \frac{a^2(1+p^2)(2+q^2+p^2-(q-p)^2)}{2 a^2 (1+p^2)\sqrt{(1+p^2)(1+q^2)}} \\ &&&= \frac{1+pq}{\sqrt{(1+p^2)(1+q^2)}} \end{align*} As required. Notice that by symmetry, \(\cos \angle TFQ = \frac{1+qp}{\sqrt{(1+q^2)(1+p^2)}} = \cos \phi\). Therefore they have the same angle and \(FT\) bisects \(PFQ\)

Solution: \begin{align*} && \tan \theta &= y/x \\ \Rightarrow && \sec^2 \theta \frac{\d \theta}{\d t} &= \frac{x \frac{\d y}{\d t} - y \frac{\d x}{\d t}}{x^2} \\ \Rightarrow && \frac{\d \theta}{\d t} &=\left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} \right) \frac{\cos^2 \theta}{x^2} \\ &&&=\left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} \right) \frac{\cos^2 \theta}{r^2 \cos^2 \theta } \\ &&&=\left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} \right) \frac{1}{r^2 } \\ && \tfrac12 \int r^2 \, \d \theta &= \tfrac12 \int \left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} \right) \d t \end{align*} \(A = (x - a \cos t, y - a \sin t), B = (x + b \cos t , y + b \sin t)\) \begin{align*} && [A] &= \tfrac12 \int_0^{2\pi} \left ((x-a \cos t) \frac{\d (y-a \sin t)}{\d t} - (y-a \sin t) \frac{\d (x-a \cos t)}{\d t} \right) \d t \\ &&&= \tfrac12 \int_0^{2\pi} \left ((x - a \cos t) \left ( \frac{\d y}{\d t} - a \cos t \right) - (y - a \sin t) \left ( \frac{\d x}{\d t} + a \sin t \right)\right) \d t \\ &&&= \tfrac12 \int_0^{2\pi} \left ( x \frac{\d y}{\d t} - y \frac{\d x}{\d t} - a \cos t \frac{\d y}{\d t}-ax \cos t +a^2 \cos^2 t + a \sin t \frac{\d x}{\d t}-y a \sin t + a^2 \sin^2 t \right) \d t \\ &&&= \tfrac12 \int_0^{2\pi} \left ( \underbrace{x \frac{\d y}{\d t} - y \frac{\d x}{\d t}}_{[P]}-a\left ((x + \frac{\d y}{\d x}) \cos t + (y - \frac{\d x}{\d t}) \sin t \right) + \underbrace{a^2}_{\pi a^2} \right) \d t \\ &&&= [P] + \pi a^2 - af \end{align*} \begin{align*} && [B] &= \tfrac12 \int_0^{2\pi} \left ((x+b \cos t) \frac{\d (y+b \sin t)}{\d t} - (y+b \sin t) \frac{\d (x+b \cos t)}{\d t} \right) \d t \\ &&&= \tfrac12 \int_0^{2\pi} \left ((x+b \cos t) (\frac{\d y}{\d t} + b \cos t) - (y+b \sin t)(\frac{\d x}{\d t} - b \sin t) \right) \d t \\ &&&= \tfrac12 \int_0^{2\pi} \left (x \frac{\d y}{\d t} - y \frac{\d x}{\d t} + b^2 + b(\cos t (x + \frac{\d y}{\d t}) +(y - \frac{\d x}{\d t})\sin t\right) \d t \\ &&&= [P] + \pi b^2 + b f \end{align*} Since \(A\) and \(B\) trace out the same area, we must have \(\pi a^2 - af = \pi b^2 + bf \Rightarrow \pi (a^2-b^2) = f(b+a) \Rightarrow f = \pi (a-b)\). In particular the area inbetween is \([A] - [P] = \pi a^2 - a \pi (a-b)\)

Solution: \begin{align*} w &= \frac{1+iz}{i+z} \\ &= \frac{1-y+ix}{x+i(1+y)} \\ &= \frac{((1-y)+ix)(x-i(1+y))}{x^2+(1+y)^2} \\ &= \frac{x(1-y)+x(1+y)}{x^2+(1+y)^2}+\frac{x^2+y^2-1}{x^2+(1+y)^2}i \\ &= \frac{2x}{x^2+(y+1)^2}+\frac{x^2+y^2-1}{x^2+(1+y)^2}i \end{align*} Therefore \(u = \frac{2x}{x^2+(y+1)^2}, v = \frac{x^2+y^2-1}{x^2+(1+y)^2}\)

1. Suppose \(z = \tan(\theta/2) = t\) then \(u = \frac{2t}{t^2+1} = \sin \theta, v = \frac{t^2-1}{t^2+1} = \cos \theta\), ie \(u+iv\) is the unit circle, where \(-\frac{\pi}{2} < \theta/2 < \frac{\pi}{2}\) or \(-\pi < \theta < \pi\), ie excluding the point \((\sin \pi, \cos \pi) = (0,1)\).
2. When \(-1 < x < 1\) we have \(-\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{4}\) ie \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), ie the lower half of the unit circle.
3. When \(x = 0, -1 < y < 1\) we have \(u = 0, v = \frac{y^2-1}{(1+y)^2}\) which is the negative imaginary axis.
4. We have \(u = \frac{2t}{t^2+4}, v = \frac{t^2}{t^2+4}\), ie \(u^2 + v^2 = v\), ie \(u^2+(v-\frac12)^2 = \frac12^2\), so a circle centre \(\frac12i\) radius \(\frac12\), missing out \((0,1)\)

Solution: \begin{align*} && {\bf r} &=\left( \e^{t}\cos t \right) {\bf i}+ \left(\e^t \sin t\right) {\bf j} \\ \Rightarrow && \dot{\bf r} &= \left( \e^{t}\cos t -\e^t \sin t\right) {\bf i}+ \left(\e^t \sin t+\e^t \cos t\right) {\bf j} \\ \Rightarrow && \mathbf{r}\cdot\dot{ \mathbf{r}} &= e^{2t}(\cos^2 t - \sin t \cos t) + e^{2t}(\sin^2 t+ \sin t \cos t) \\ &&&= e^{2t} (\cos^2 t + \sin ^2 t)\\ &&&= e^{2t} \\ \\ && | {\bf r}| &= e^{t} \\ && |{\bf \dot{r}}| &= e^t \sqrt{(\cos t - \sin t)^2 + (\sin t + \cos t)^2} \\ &&&= e^t \sqrt{2 \cos^2 t + 2 \sin^2 t} \\ &&&= \sqrt{2} e^t \\ \\ \Rightarrow && \frac{\mathbf{r}\cdot\dot{ \mathbf{r}}}{ |{\bf {r}}| |{\bf \dot{r}}|} &= \frac{e^{2t}}{\sqrt{2}e^te^t} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} Therefore the angle between the velocity and displacement is \(\frac{\pi}{4}\). \begin{align*} && \ddot{\bf{r}} &= \left( \e^{t}(\cos t - \sin t) - \e^t (\sin t + \cos t)\right) {\bf i}+ \left(\e^t (\sin t + \cos t) + \e^t(\cos t - \sin t)\right) {\bf j} \\ &&&= \left ( -2\e^{t} \sin t \right) {\bf i}+ \left ( 2\e^{t} \cos t \right) {\bf j} \\ \Rightarrow && {\bf r} \cdot \ddot{\bf{r}} &= 2e^{2t} \left ( -\sin t \cos t + \sin t \cos t \right) \\ &&&= 0 \end{align*} Therefore the acceleration is perpendicular.

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