2 problems found
Let \(P\) be the point on the curve \(y=ax^2+bx+c\) (where \(a\) is non-zero) at which the gradient is \(m\). Show that the equation of the tangent at \(P\) is \[ y-mx=c-\frac{(m-b)^2}{4a}\;. \] Show that the curves \(y=a_1 x^2+b_1 x+c_1\) and \(y=a_2 x^2+b_2 x+c_2\) (where \(a_1\) and \(a_2\) are non-zero) have a common tangent with gradient \(m\) if and only if \[ (a_2 -a_1 )m^2 + 2(a_1 b_2-a_2 b_1)m + 4a_1 a_2(c_2-c_1)+ a_2 b_1^2-a_1 b_2 ^2=0\;. \] Show that, in the case \(a_1 \ne a_2 \,\), the two curves have exactly one common tangent if and only if they touch each other. In the case \(a_1 =a_2\,\), find a necessary and sufficient condition for the two curves to have exactly one common tangent.
Solution: \begin{align*} && y' &= 2ax+b \\ \Rightarrow && m &= 2ax_t+b \\ \Rightarrow && x_t &= \frac{m-b}{2a} \end{align*} Therefore we must have \begin{align*} mx_t &= 2ax_t^2+bx_t \\ y - mx &= ax_t^2+bx_t+c - mx_t \\ &= ax_t^2+bx_t+c - (2ax_t^2+bx_t) \\ &= c - ax_t^2 \\ &= c-a\left (\frac{m-b}{2a} \right)^2 \\ &= c - \frac{(m-b)^2}{4a} \end{align*} They will have a common tangent if and only if the constant terms are equal, ie \begin{align*} && c_1 - \frac{(m-b_1)^2}{4a_1} &= c_2 - \frac{(m-b_2)^2}{4a_2} \\ \Leftrightarrow && (c_1-c_2) &= \frac{(m-b_1)^2}{4a_1} -\frac{(m-b_2)^2}{4a_2} \\ \Leftrightarrow && 4a_1a_2(c_1-c_2) &= a_2(m-b_1)^2-a_1(m-b_2)^2 \\ &&&= (a_2-a_1)m^2+2(a_1b_2-a_2b_1)m+a_2b_1^2-a_1b_2^2 \end{align*} as required. Treating this as a polynomial in \(m\), we can see that the two curves will have exactly one common tangent iff \(\Delta = 0\), ie: \begin{align*} && 0 &= \Delta \\ &&&= (2(a_1b_2-a_2b_1))^2 - 4 (a_2-a_1)(4a_1 a_2(c_2-c_1)+ a_2 b_1^2-a_1 b_2 ^2) \\ &&&= 4a_1^2b_2^2-8a_1a_2b_1b_2+4a_2b_1^2 - 4a_2^2b_1^2-4a_1^2b_2^2 + 4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\ &&&=-8a_1a_2b_1b_2+4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\ &&&=a_1a_2(4(b_1-b_2)^2-16(a_2-a_1)(c_2-c_1)) \\ &&&= 4a_1a_2((b_2-b_1)^2 - 4(a_2-a_1)(c_2-c_1) \end{align*} But this is just the discriminant of the difference, ie equivalent to the two parabolas just touching. (Assuming \(a_1-a_2 \neq 0\) and we do end up with a quadratic). If \(a_1 = a_2 = a\) then we need exactly one solution to \(2a(b_1-b_2)m +4a^2(c_2-c_1)+a(b_1^2-b_2^2) = 0\), ie \(b_1 \neq b_2\).
Find the two solutions of the differential equation \[ \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}=4y \] which pass through the point \((a,b^{2}),\) where \(b\neq0.\) Find two distinct points \((a_{1},1)\) and \((a_{2},1)\) such that one of the solutions through each of them also passes through the origin. Show that the graphs of these two solutions coincide and sketch their common graph, together with the other solutions through \((a_{1},1)\) and \((a_{2},1)\). Now sketch sufficient members of the family of solutions (for varying \(a\) and \(b\)) to indicate the general behaviour. Use your sketch to identify a common tangent, and comment briefly on its relevance to the differential equation.