- 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
- Show that the vector $T = y\,\mathbf{i} + 2c\,\mathbf{j}$ is tangent to the parabola $y^2 = 4cx$ at the point $(x, y),$ and that the vector $N = 2c\,\mathbf{i} - y\,\mathbf{j}$ is perpendicular to $T.$ [Hint: Write a vector equation for the parabola, using $y$ as a parameter.]
- To find the vector equation for the parabola using $y$ as a parameter, we rewrite the equation $y^2 = 4cx$ as $\frac{y^2}{4c} = x,$ which gives us: \begin{align*} \mathbf{r}(y) &= \frac{y^2}{4c}\,\mathbf{i} + y\,\mathbf{j} \end{align*}
- $N$ and $T$ are perpendicular since $N \cdot T = 0.$ To show that $T$ is tangent to the parabola $y^2 = 4cx$ at $(x, y),$ it will suffice to show that $T$ is parallel to the derivative of the curve parameterized by $y.$ Writing the parabola as a vector equation, using $y$ as a parameter, we get: \begin{align*} \\ \mathbf{r}(y) &= \frac{y^2}{4c}\,\mathbf{i} + y\,\mathbf{j} \end{align*} Its derivative is then: \begin{align*} \mathbf{r}'(y) &= \frac{y}{2c}\,\mathbf{i} + \mathbf{j} \end{align*} But we can see that that $T$ is parallel to $\mathbf{r}'(y),$ with $T = 2c\,\mathbf{r}'(y).$ Thus, $T$ is tangent to the parabola at $(x, y).\,\blacksquare$