- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
13.25. Miscellaneous review exercises on conic sections
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(a) A chord of length $8|c|$ is drawn perpendicular to the axis of the parabola $y^2 = 4cx.$ Let $P$ and $Q$ be the points where the chord meets the parabola. Show that the vector from $O$ to $P$ is perpendicular to that from $O$ to $Q.$
(b) The chord of a parabola drawn through the focus and parallel to the directrix is called the latus rectum. Show first that the length of the latus rectum is twice the distance from the focus to the directrix, and then show that the tangents to the parabola at both ends of the latus rectum intersect the axis of the parabola on the directrix.
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(a) Recall that a parabola is symmetric about its axis (that is, the line defined by the focus and vertex). What can we deduce from this symmetry regarding the chord from $P$ to $Q$?
(b) What are the points at which the latus rectum intersect with the parabola? What is the derivative of the function at those two points?
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(a) The parabola described by the Cartesian equation $y^2 = 4cx$ is symmetric about the $x$-axis, thus the chord of length $8|c|$ drawn perpendicular to the $x$ axis can be described as two chords of length $4|c|$ drawn from the $x$-axis to the points $P = (4c, 4c)$ and $Q = (4c, -4c).$ Taking the dot product of these two points (that, is, vectors from $O$ to $P$ and $Q,$ respectively), we get $0.$ Thus, the two vectors are perpendicular. $\blacksquare$
(b) First, we note that the Cartesian equation $y^2 = 4cx$ describes a parabola with focus $F = (c, 0)$ and vertex at $O,$ thus putting the directrix line at $x = -c,$ making the axis of the parabola the $x$ axis.
Then, we find the points on the parabola where $x = c,$ giving us $(c, 2c)$ and $(c, -2c).$ This means that the length of the latus rectum is $4c.$ But we know that the distance from $(c, 0)$ to $(-c, 0)$ is $2c,$ thus the latus rectum has length of twice the distance from the focus to the directrix. To find the slope of the lines tangent to the parabola at $(c, 2c)$ and $(c, -2c),$ we take the derivative with respect to $y:$ $$\frac{d}{dy}\left(\frac{y^2}{4c}\right) = \frac{y}{2c}$$Evaluating this at $y=2c$ and $y=-2c,$ we get slopes $1$ and $-1,$ respectively, meaning that the line tangent to the parabola at $(c, 2c)$ has equation $x = y + b.$ Evaluating the line at $(c, 2c),$ we find that $b,$ the $x$-intercept, is $-c,$ meaning that the line intercepts the $x$ axis through the directrix line. Doing this at $(c, -2c)$ gives a tangent line with equation $x = b - y,$ evaluating at $(c, -2c)$ gives $b = -c,$ meaning that this tangent line also intercepts the $x$ axis through the directrix line. This completes the proof.
