- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.21 Miscellaneous Review Exercises
- Prove that an equation of the line of slope $m$ that is tangent to the parabola $y^2 = 4cx$ can be written in the form $y = mx + c/m.$ What are the coordinates of the point of contact?
- Note that slope $m$ is a constant. Take the derivative of the parabola with respect to $x$ and use slope $m$ to find the $y$-coordinate of the point of contact. From here, use $m$ and the point of contact to find the $y$-intercept.
- Taking the derivative of the parabola with respect to $x,$ we get: \begin{align*} \\ 2y\frac{dy}{dx} &= 4c \end{align*} Setting $dy/dx = m$ and rearranging terms we find that, at the point of contact, $y = 2c/m.$ Then, $y^2 = 4c^2/m^2 = 4cx,$ which means that the point of contact is: \begin{align*} \\ \left(\frac{c}{m^2}, \frac{2c}{m}\right) \end{align*} Putting this point into slope-intercept form, we get: \begin{align*} \\ y &= mx + b \Rightarrow \frac{2c}{m} = \frac{c}{m} + b \end{align*} Solving for $b,$ we see that the equation for the line of slope $m$ tangent to the curve $y^2 = 4cx$ is \begin{align*} \\ y &= mx + \frac{c}{m} \quad \blacksquare \end{align*}