Problems

Solution:

+ - ↺

1. \(\,\) \begin{align*} \bf{COE}: && \frac12 m \cdot 5ag &= mg\cdot 2a + \frac12 m v^2 \\ \Rightarrow && v^2 &= ag \\ && v &= \sqrt{ag} \end{align*} If it just clears the second wall, we must have: \begin{align*} && 0 &= \sqrt{ag} \sin \theta t - \frac12 gt^2 \\ \Rightarrow && t &= \frac{2\sqrt{ag}\sin \theta}{g} \\ && a &= \sqrt{ag} \cos \theta t \\ &&&=\sqrt{ag} \cos \theta \frac{2\sqrt{ag}\sin \theta}{g} \\ &&&= a \sin 2 \theta \\ \Rightarrow && \theta &= 45^{\circ} \end{align*} Imagine firing the particle backwards from the top of the wall at \(45^\circ\) then \begin{align*} && -2a &= \sqrt{ag}\cdot \left ( -\frac1{\sqrt{2}} \right) t - \frac12 g t^2 \\ \Rightarrow && 0 &= gt^2+\sqrt{2ag} t -4a \\ &&&= (\sqrt{g}t -\sqrt{2} \sqrt{a})(\sqrt{g}t +2\sqrt{2} \sqrt{a}) \\ \Rightarrow && t &= \sqrt{\frac{2a}{g}} \\ \Rightarrow && s &= \left ( -\frac1{\sqrt{2}} \right) \sqrt{ag} \sqrt{\frac{2a}{g}} \\ &&&= -a \end{align*} Therefore the \(A\) is \(a\) from the wall.
2. When it passes over the first wall, \begin{align*} \bf{COE}: && \frac52amg &= (2a+h)mg + \frac12 m v^2 \\ \Rightarrow && v^2 &= (a-2h)g \end{align*} Now imagine firing a particle with this speed in any direction. The question is asking whether we can ever travel \(2a\) without descending more than \(h\). \begin{align*} && a &= \sqrt{(a-2h)g} \cos \beta t \\ \Rightarrow && t &= \frac{a}{\sqrt{(a-2h)g} \cos \beta}\\ && -h &= \sqrt{(a-2h)g} \sin \beta t - \frac12 g t^2 \\ &&&= a \tan \beta - \frac12 \frac{a^2}{(a-2h)} \sec^2 \beta \\ &&&= a \tan \beta - \frac{a^2}{2(a-2h)}(1+ \tan^2 \beta )\\ \Rightarrow && 0 &= \frac{a^2}{2(a-2h)} \tan^2 \beta-a \tan \beta + \frac{a^2-2ah+4h^2}{2(a-2h)} \\ && \Delta &= a^2 - \frac{a^2}{a-2h} \frac{a^2-2ah+4h^2}{a-2h} \\ &&&= \frac{a^2}{(a-2h)^2}\left ( a^2-4ah+4h^2-a^2+2ah-4h^2\right) \\ &&&= \frac{a^2}{(a-2h)^2}\left ( -2ah\right) < 0 \\ \end{align*} So there are no solutions if \(h > 0\)

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(X+Y=r) &= \sum_{k=0}^r \mathbb{P}(X = k, Y = r-k) \\ &&&= \sum_{k=0}^r \mathbb{P}(X = k)\mathbb{P}( Y = r-k) \\ &&&= \sum_{k=0}^r \frac{e^{-\lambda T} (\lambda T)^k}{k!}\frac{e^{-\mu T} (\mu T)^{r-k}}{(r-k)!}\\ &&&= \frac{e^{-(\mu+\lambda)T}}{r!}\sum_{k=0}^r \binom{r}{k}(\lambda T)^k (\mu T)^{r-k}\\ &&&= \frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^r}{r!} \end{align*} Therefore \(X+Y \sim Po \left ( (\mu+\lambda)T \right)\)
2. \(\,\) \begin{align*} && \mathbb{P}(X = r | X+Y = k) &= \frac{\mathbb{P}(X=r, Y = k-r)}{\mathbb{P}(X+Y=k)} \\ &&&= \frac{\frac{e^{-\lambda T} (\lambda T)^r}{r!}\frac{e^{-\mu T} (\mu T)^{k-r}}{(k-r)!}}{\frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^k}{k!}} \\ &&&= \binom{k}{r} \left ( \frac{\lambda}{\lambda + \mu} \right)^r \left ( \frac{\mu}{\lambda + \mu} \right)^{k-r} \end{align*} Therefore \(X|X+Y=k \sim B(k, \frac{\lambda}{\lambda + \mu})\)
3. \(P(X=1|X+Y = 1) = \frac{\lambda}{\lambda + \mu}\)
4. Let \(X_1, Y_1\) be the time to the first fish are caught by Adam and Eve, then \begin{align*} && \mathbb{P}(X_1, Y_1 > t) &= \mathbb{P}(X_1> t) \mathbb{P}( Y_1 > t) \\ &&&= e^{-\lambda t}e^{-\mu t} \\ &&&= e^{-(\lambda+\mu)t} \\ \Rightarrow && f_{\max(X_1,Y_1)}(t) &= (\lambda+\mu)e^{-(\lambda+\mu)} \end{align*} Therefore the expected time is \(\frac1{\mu+\lambda}\)

Solution:

1. The probability they pull the key out on the \(k\)th attempt will be \(\left ( \frac{n-1}{n} \right)^{k-1} \frac1n\), so we want: \begin{align*} \E[G_1] &= \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \frac1n \\ &= \frac{1}n \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \\ &= \frac1n \frac{1}{\left (1 - \frac{n-1}{n} \right)^2} \\ &= \frac{1}{n} \frac{n^2}{1^2} = n \end{align*}
2. In version 2, the probability the correct key comes out at the \(k\)th attempt is \(\frac1n\) (assume we take out all the keys, then the correct key is equally likely to appear in all of the space). Therefore \(\E[G_2] = \frac1n (1 + 2 + \cdots + n) = \frac{n+1}{2}\)
3. The probability the key comes out on the correct attempt is: \begin{align*} && \mathbb{P}(G_3 = k) &= \frac{n-1}{n} \cdot \frac{n}{n+1} \cdot \frac{n+1}{n+2} \cdots \frac{n+k-3}{n+k-2} \cdot \frac{1}{n+k-1} \\ &&&= \frac{n-1}{(n+k-2)(n+k-1)} \\ \\ &&k \cdot \mathbb{P}(G_3 = k) &= \frac{k(n-1)}{(n+k-2)(n+k-1)} \\ &&&= \frac{(n-1)(2-n)}{n+k-2} + \frac{(n-1)^2}{n+k-1} \\ &&&= \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n-k+2} \\ \Rightarrow && \E[G_3] &= \sum_{k=1}^{\infty} k \cdot \mathbb{P}(G_3 = k) \\ &&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n+k-2} \right) \\ &&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} \right) +\underbrace{\sum_{k=1}^{\infty} \frac{n-1}{n-k+2}}_{\to \infty} \\ \end{align*}

Solution: \begin{align*} \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) &= \frac{r+1}{r} \l \frac{r!(n-1)!}{(n+r-1)!} - \frac{r!n!}{(n+r)!} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \l 1 - \frac{n}{n+r} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \frac{r}{n+r} \\ &= \frac{(r+1)!n!}{(n+r)!} \\ &= \frac{1}{^{n+r}\C_{r+1}} \end{align*} \begin{align*} \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} &= \sum_{n=1}^{\infty} \l \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) \r \\ &= \frac{r+1}{r} \sum_{n=1}^{\infty} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \sum_{n=1}^{N} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \l \frac{1}{^{1+r-1}\C_{r}} - \frac{1}{^{N+r}\C_{r}}\r \\ &= \frac{r+1}{r} \frac{1}{^{1+r-1}\C_{r}} \tag{since \(\frac{1}{^{N+r}\C_{r}} \to 0\)} \\ &= \frac{r+1}{r} \end{align*} When \(r = 2\), we have: \begin{align*} && \frac{3}{2} &= \sum_{n=1}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &=\frac{1}{^{1+2}\C_{3}} + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &= 1 + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ \Rightarrow && \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} &= \frac12 \end{align*} \begin{align*} \frac{1}{^{n+1}\C_{3}} &= \frac{3!}{(n+1)n(n-1)} \\ &= \frac{3!}{n^3-n} \\ &> \frac{3!}{n^3} \end{align*} \begin{align*} \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &= \frac{5!}{(n+1)n(n-1)} - \frac{5!}{(n+2)(n+1)n(n-1)(n-2)} \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l 1- \frac{1}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l \frac{n^2-5}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2(n^2-5)}{(n^2-1)(n^2-4)} \\ &< \frac{5!}{n^3} \end{align*} Since \(k(k-5) < (k-1)(k-4) \Leftrightarrow 0 < 4\), this only makes sense if \(n \geq 3\) \begin{align*} &&\frac{3!}{n^3} &< \frac{1}{^{n+1}\C_{3}} \tag{if \(n \geq 3\)} \\ \Rightarrow &&\sum_{n=3}^\infty \frac{3!}{n^3} &< \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{3!}{n^3} &< \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \sum_{n=2}^\infty \frac{1}{^{n+2}\C_{2+1}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \frac{1}{2} = \frac{29}{4} \\ \Rightarrow && \sum_{n=1}^\infty \frac{1}{n^3} &< \frac{29}{24} = \frac{116}{96} \\ \end{align*} \begin{align*} && \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &< \frac{5!}{n^3} \\ \Rightarrow && \sum_{n=3}^\infty \l \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} \r &< \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{20}{^{n+1}\C_3} - \sum_{n=3}^\infty \frac{1}{^{n+2}\C_{5}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=2}^\infty \frac{20}{^{n+2}\C_{2+1}} - \sum_{n=1}^\infty \frac{1}{^{n+4}\C_{4+1}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \frac{20}{2} - \frac{4+1}{4} &< \sum_{n=1}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{115}{96} &< \sum_{n=1}^\infty \frac{1}{n^3} \\ \end{align*}

Solution:

1. We can map \(a \mapsto 0\), rotate the whole plane, then shift the plane back to \(a\), so \(z \mapsto (z-a) \mapsto e^{i \theta}(z-a) \mapsto a + e^{i \theta}(z-a) = e^{i \theta}z + a(1 - e^{i\theta})\)
2. \(z \mapsto e^{i \theta}z + a(1 - e^{i\theta}) \mapsto e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi})\) \begin{align*} e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi}) &= e^{i(\phi + \theta)}z + ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \\ \end{align*} Therefore this is rotation by angle \(\phi + \theta\) and about \begin{align*} \frac{ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi})}{1 - e^{i(\phi + \theta)}} &= \frac{e^{-i\frac{(\phi + \theta)}{2}} \l ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{\l ae^{i\frac{\phi-\theta}{2}} - ae^{i \frac{(\theta + \phi)}{2}} + b(e^{-i\frac{(\phi + \theta)}{2}} -e^{i\frac{(\phi - \theta)}{2}}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{ae^{i\frac{\phi}{2}} 2i\sin(\frac{\theta}{2}) + be^{-i\frac{\theta}{2}}2i\sin\frac{\phi}{2} }{2i \sin(\frac{\phi + \theta}2)} \\ \end{align*} as required. If \(\phi + \theta = 2\pi\), then \(z \mapsto z + (b-a)(1 - e^{i\phi})\) which is a translation.
3. If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \begin{align*} && a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi &= b\,\e^{ {\mathrm i}\theta/2}\sin \tfrac12\phi + a\,\e^{-{\mathrm i} \phi/2}\sin \tfrac12 \theta \\ && a\,(\e^{ {\mathrm i}\phi/2}-\e^{-{\mathrm i}\phi/2})\sin \tfrac12\theta + b\,(\e^{-{\mathrm i} \theta/2}-\e^{+{\mathrm i} \theta/2})\sin \tfrac12 \phi &= 0 \\ && a \sin \frac{\phi}{2} \sin \frac{\theta}{2}-b \sin \frac{\theta}{2} \sin \frac{\phi}{2} &= 0 \\ \Leftrightarrow && a = b \text{ or } \sin \frac{\theta}{2} = 0 \text{ or } \sin \frac{\phi}{2} = 0 \\ \Leftrightarrow && a = b \text{ or } \theta = 0 \text{ or } \phi = 0 \\ \end{align*} If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \(b = a\) or \(e^{i\phi} = e^{i \theta}\) ie rotation through the same angle.

Solution: \begin{align*} A &= -(\alpha \beta + \gamma\delta + \alpha\gamma+\beta\delta+\alpha \delta + \beta\gamma) \\ &= -q \end{align*}

1. The corresponding cubic equation is: \begin{align*} && 0 &= y^3 - 3y^2-40y+(120-36) \\ &&&= y^3 -3y^2 - 40y + 84 \\ &&&= (y-7)(y-2)(y+6) \end{align*} Therefore \(\alpha\beta + \gamma \delta = 7\)
2. \begin{align*}(\alpha+\beta)(\gamma+\delta) &= \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta \\ &= 3 -(\alpha\beta + \gamma\delta) \\ &=3-7 = -4 \end{align*} Let \(\alpha\beta\) and \(\gamma\delta\) be the roots of a quadratic; then the quadratic will be \(t^2-7t+10 = 0 \Rightarrow t = 2,5\) so \(\alpha\beta = 5\)
3. \(\alpha\beta = 5, \gamma\delta = 2\) Consider the quadratic with roots \(\alpha+\beta\) and \(\gamma+\delta\), then \(t^2-4 = 0 \Rightarrow t = \pm 2\). Suppose \(\alpha+\beta = 2, \gamma+\delta=-2\) then \(\alpha, \beta = 1 \pm 2i, \gamma,\delta = -1 \pm i\) \(\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = -6 \neq 6\) Suppose \(\alpha+\beta = -2, \gamma+\delta=2\) then \(\alpha, \beta = -1 \pm 2i, \gamma,\delta = 1 \pm i\) \(\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = 6\), therefore these are there roots. (In some order): \(1 \pm i, -1 \pm 2i\)

Solution:

1. \begin{align*} && a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} &= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} \\ &&&= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \frac{\ln f(x)}{\ln a} \, dx} \\ &&&= e^{ \frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \\ &&&= F(y) \end{align*}
2. \(\,\) \begin{align*} && H(y) &= e^{\frac1y \int_0^y \ln h(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln (f(x)g(x))\d x} \\ &&&= e^{\frac1y \int_0^y \left ( \ln f(x)+\ln g(x) \right)\d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x +\frac1y \int_0^y \ln g(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x }e^{\frac1y \int_0^y \ln g(x) \d x} \\ &&&= F(y)G(y) \end{align*}
3. Suppose \(f(x) = b^x\), then \begin{align*} && F(y) &= b^{\frac1y \int_0^y \log_b f(x) \d x} \\ &&&= b^{\frac1y \int_0^y x \d x}\\ &&&= b^{\frac1y \frac{y^2}{2}} \\ &&&= b^{\frac{y}2} = \sqrt{b^y} \end{align*}
4. Suppose the geometric mean of \(f(x)\) is \(\sqrt{f(y)}\) then the geometric mean of \(f(x)^2\) is \(f(y)\) by the the second part.

Solution: \((x, y) = (f(\theta)\cos(\theta), f(\theta)\sin(\theta))\) so \begin{align*} \frac{dy}{d\theta} &= -f(\theta)\sin(\theta) + f'(\theta)\cos(\theta) \\ \frac{dx}{d\theta} &= f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) \\ \frac{dy}{dx} &= \frac{-f(\theta)\sin(\theta) + f'(\theta)\cos(\theta)}{f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) } \\ &= \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \end{align*} If the curves meet at right angles then the product of their gradients is \(-1\), ie \begin{align*} \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \cdot \frac{-g(\theta)\tan(\theta) + g'(\theta)}{g(\theta) + g'(\theta)\tan(\theta) } &= -1 \\ f(\theta)g(\theta)\tan^2 \theta - f(\theta)g'(\theta)\tan \theta - f'(\theta)g(\theta)\tan \theta + f'(\theta)g'(\theta) &= \\ \quad - \l f(\theta)g(\theta) + f(\theta)g'(\theta)\tan(\theta) + f'(\theta)g(\theta)\tan(\theta) + f'(\theta)g'(\theta)\tan^2 \theta \r \\ \tan^2\theta \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r + f'(\theta)g'(\theta) + f(\theta)g(\theta) &= 0 \\ (\tan^2\theta + 1) \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r &= 0 \\ f(\theta)g(\theta) + f'(\theta)g'(\theta) &= 0 \end{align*} \(g(\theta) = a(1+\sin\theta), g'(\theta) = a\cos\theta\) Therefore \(f'(\theta)a\cos \theta+f(\theta)a(1+\sin(\theta)) = 0\) \begin{align*} && \frac{f'(\theta)}{f(\theta)} &= -\sec(\theta) - \tan(\theta) \\ \Rightarrow && \ln(f(\theta)) &= -\ln |\tan(\theta) + \sec(\theta)| + \ln |\cos(\theta)| + C \\ \Rightarrow && f(\theta) &= A \frac{\cos \theta}{\tan \theta + \sec \theta} \\ &&&= A \frac{\cos^2 \theta}{\sin \theta + 1} \\ &&&= A \frac{1-\sin^2 \theta}{\sin \theta + 1} \\ &&&= A (1-\sin \theta) \end{align*} When \(\theta = -\frac12 \pi, r = 4\), so \(A = 2\).

+ - ↺

Solution:

1. The tangent has equation: \begin{align*} && 0 &= \frac{Xx}{a^2} + \frac{Yy}{b^2} -1 \\ \Rightarrow &&&= \frac{Xa(1-t^2)}{a^2(1+t^2)} + \frac{Y2bt}{b^2(1+t^2)} - 1 \\ \Rightarrow &&0&= Xb(1-t^2) + Y2at - ab(1+t^2)\\ &&&= -(b(a+X)t^2 -2aYt +b(a-X)) \\ \Rightarrow && 0 &= (a+X)bt^2-2aYt+b(a-X) \\ \\ && 0 <\Delta &= 4a^2Y^2 - 4(a+X)b(a-X)b \\ &&&= 4(a^2Y^2-b^2(a^2-X^2)) \\ \Leftrightarrow && a^2Y^2 &> (a^2-X^2)b^2 \end{align*} Therefore there are two roots to the quadratic, ie two values of the parameter \(t\) which works. The condition is equivalent to \(\frac{X^2}{a^2} + \frac{Y^2}{b^2} > 1\). ie from any point outside the ellipse there are two tangent lies.
   

   + - ↺
   Clearly there are two tangents when \(X = \pm a\) (except \((X,Y) = (\pm a, 0)\).
2. We must have \(p\) and \(q\) are roots of \(0 = (a+X)bt^2-2aYt+b(a-X)\), ie \(pq = \frac{b(a-X)}{(a+X)b} \Rightarrow (a+X)pq = a-X\). Similarly \(p+q = \frac{2aY}{(a+X)b}\) Given that the tangents meet the \(y\)-axis at \((0, y_i)\) we must have \(abt^2-2ay_it + ab = 0\), so \begin{align*} && 0 &= abp^2-2ay_1p + ab \\ && 0 &= abq^2-2ay_2q + ab \\ \Rightarrow && y_1 &= \frac{ab(p^2+1)}{2ap} \\ && y_2 &= \frac{ab(q^2+1)}{2aq} \\ \Rightarrow && 2b &= \frac{ab(p^2+1)}{2ap} +\frac{ab(q^2+1)}{2aq} \\ &&&= \frac{ab(pq(p+q)+p+q)}{2apq} \\ \Rightarrow && 4pq &= pq(p+q)+p+q \\ \Rightarrow && 4 \frac{b(a-X)}{(a+X)b} &= \frac{2aY}{(a+X)b} \left ( \frac{b(a-X)}{(a+X)b} + 1 \right) \\ && &= \frac{2aY}{(a+X)b} \frac{2ab}{(a+X)b} \\ \Rightarrow && 4b^2(a^2-X^2) &= 4a^2bY \\ \Rightarrow && 1 &= \frac{Y}{b} + \frac{X^2}{a^2} \end{align*} as required.