Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• (a) Solve Exercise 3 for the parabola $(y - y_0)^2 = 4c(x - x_0).$ (b) Solve Exercise 3 for the parabola $x^2 = 4cy$ and, more generally, for the parabola $(x - x_0)^2 = 4c(y - y_0).$
• (a) What happens to the point of contact when $x$ and $y$ are translated by $x_0$ and $y_0,$ respectively? (b) What happens to our graphs and their point of contact when the axes are switched, but $m$ stays the same?
• (a) As in Exercise 3, we take the derivative of the equation with respect to $x$ and set $dy/dx = m$ to find the point of contact: \begin{align*} \\ 2(y - y_0)\frac{dy}{dx} = 4c &\Rightarrow y = \frac{2c}{m} + y_0 \\ \\ (y - y_0)^2 = 4c(x - x_0) &\Rightarrow x = \frac{1}{4c}(y - y_0)^2 + x_0 \\ \\ &\Rightarrow x = \frac{c}{m^2} + x_0 \end{align*} As we can see, the point of contact is that from Exercise 3, translated by $(x_0, y_0).$ Putting the point of contact into slope-intercept form, we find the $y$-intercept: \begin{align*} \\ y = mx + b &\Rightarrow \frac{2c}{m} + y_0 = \frac{c}{m} + mx_0 + b \\ \\ &\Rightarrow b = \frac{c}{m} + y_0 - mx_0 \end{align*} The equation of the line of slope $m$ tangent to the parabola $(y - y_0)^2 = 4c(x - x_0)$ is then: \begin{align*} \\ y = mx + \frac{c}{m} + y_0 - mx_0 \end{align*} Rearranging terms gives us the equation of the line in translated point-slope form: \begin{align*} \\ y - y_0 &= m(x - x_0) + \frac{c}{m} \quad \blacksquare \end{align*} (b) We first take the derivative of the equation $x^2 = 4cy$ with respect to $x$ and set $dy/dx = m$ to find the point of contact: \begin{align*} \\ 2x = 4c\frac{dy}{dx} &\Rightarrow x = 2cm \\ \\ x^2 = 4cy &\Rightarrow y = cm^2 \end{align*} Putting this point into slope-intercept form, we get \begin{align*} \\ y &= mx + b \Rightarrow cm^2 = 2cm^2 + b \end{align*} Making the $y$-intercept $b = -cm^2.$ The line of slope $m$ tangent to the parabola $x^2 = 4cy$ then has the equation: \begin{align*} \\ y &= mx - cm^2 \end{align*} In the general case of a parabola translated by $(x_0, y_0),$ we translate the point of contact accordingly, giving us: \begin{align*} \\ (x, y) &= \left(2cm + x_0, cm^2 + y_0\right) \end{align*} Putting the point of contact into slope-intercept form, we get: \begin{align*} \\ cm^2 + y_0 &= 2cm^2 + mx_0 + b \end{align*} Solving for the $y$-intercept $b,$ we get $b = y_0 - mx_0 -cm^2,$ making the equation of the line of slope $m$ tangent to the parabola $(x - x_0)^2 = 4c(y - y_0):$ \begin{align*} \\ y &= mx + y_0 - mx_0 - cm^2 \end{align*} Rearranging terms gives us the equation in translated point-slope form: \begin{align*} \\ y - y_0 &= m(x - x_0) - cm^2 \quad \blacksquare \end{align*}