Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• Solve Exercise 5 for each of the parabolas described in Exercise 4.
• Use the results of Exercise 4 to give the equations and points of contact $(x_1, y_1)$ for the three tangent lines.
• (a) Following the result of Exercise 4(a), we know that the equation of the line of slope $m$ tangent to the parabola with equation $(y - y_0)^2 = 4c(x - x_0)$ is: \begin{align*} y - y_0 &= m(x - x_0) + \frac{c}{m} \end{align*} with point of contact: \begin{align*} \\ (x_1, y_1) &= \left(\frac{c}{m^2} + x_0, \frac{2c}{m} + y_0\right) \end{align*} In other words, \begin{align*} \left(\frac{c}{m^2}, \frac{2c}{m}\right) &= (x_1 - x_0, y_1 - y_0) \end{align*} We can use this to express slope $m$ in terms of $y_1 - y_0,$ with $m = 2c/(y_1 - y_0),$ for $y_1 \neq y_0.$ We can also express $c/m$ in terms of $(x_1 - x_0)$ and $(y_1 - y_0)$ as: \begin{align*} \frac{c}{m} &= 2c\frac{x_1 - x_0}{y_1 - y_0} \end{align*} for $y_1 \neq y_0.$ Then, the translated point-slope equation of the tangent line becomes: \begin{align*} \\ (y - y_0) &= \frac{2c}{y_1 - y_0}\left(x + x_1 - 2x_0\right) \end{align*} Multiplying both sides by $y_1 - y_0$ gives us: \begin{align*} \\ (y - y_0)(y_1 - y_0) &= 2c\left(x + x_1 - 2x_0\right) \quad \blacksquare \end{align*} (b) Following the result of Exercise 4(b), we know that the equation of the line with slope $m,$ tangent to the parabola $x^2 = 4cy,$ is \begin{align*} \\ y = mx - cm^2 \end{align*} with point of contact $(x_1, y_1) = (2cm, cm^2).$ Then, expressing $m$ and $b$ in terms of $x_1$ and $y_1,$ we get: \begin{align*} \\ y &= \frac{y_1}{x_1}(2x - x_1) \end{align*} for $x_1 \neq 0.$ Multiplying both sides by $x_1,$ we get: \begin{align*} x_1y &= 2y_1x - x_1y_1 \quad \blacksquare \end{align*} (c) Following the result of Exercise 4(b), we know that the equation for the line with slope $m,$ tangent to the parabola $(x - x_0)^2 = 4c(y - y_0)$ is: \begin{align*} \\ y - y_0 = m(x - x_0) - cm^2 \end{align*} with point of contact being: \begin{align*} (x_1, y_1) &= \left(2cm + x_0, cm^2 + y_0\right) \end{align*} Written another way, \begin{align*} \\ \left(2cm, cm^2\right) &= \left(x_1 - x_0, y_1 - y_0\right) \end{align*} Then, we can express $m$ and $b$ in terms of $x_1 - x_0$ and $y_1 - y_0$ to give us: \begin{align*} \\ y - y_0 = \frac{y_1 - y_0}{x_1 - x_0}\left[2(x - x_0) - (x_1 - x_0)\right] \end{align*} for $x_1 \neq x_0.$ Multiplying both sides by $(x_1 - x_0)$ gives us \begin{align*} \\ (y - y_0)(x_1 - x_0) = 2(y_1 - y_0)(x - x_0) - (x_1 - x_0)(y_1 - y_0) \quad \blacksquare \end{align*}