How to find the length of the focal chord that makes an angle $\theta$ with the axis of parabola $y^2=4ax$?

Question

A focal chord of $y^2=4ax$ makes an angle $\theta$ with the axis of the parabola. How do I find the length of the chord?

I tried using the parametric equation but couldn't go further.

Answer 1

Some hints:

• What are the coordinates of the focus of the parabola?
• What is the equation of the line that makes an angle $\theta$ with the $y$-axis and goes through the focus? (This line contains the focal chord given in the question.)
• At what two points does this line intersect the parabola?
• Finally, what is the distance between these two points?

Answer 2

Consider a parabola with focus $F$ and focal chord $\overline{PQ}$ making a non-obtuse angle $\theta$ with the axis. Drop perpendiculars from $F$, $P$, $Q$ to $F'$, $P'$, $Q'$ on the directrix; and "raise" a perpendicular from $F$ to $M$ on the directrix.

By the focus-directrix definition of the parabola, $\overline{FP}\cong\overline{PP'}$ and $\overline{FQ}\cong\overline{QQ'}$. We conclude that $\square FPP'M$ and $\square FQQ'M$ are right-angled kites, with $\overline{MF}\cong\overline{MP'}\cong\overline{MQ'}$ and $\angle FMF'=\theta$. Calculating $|P'Q'|$ in two ways, we have

$$|PQ|\sin\theta = 2\,|FF'|\csc\theta \qquad\to\qquad |PQ|=2\,|FF'|\csc^2\theta \tag{$\star$}$$

Note that focus-directrix distance $|FF'|$ is twice the focus-vertex distance, which is represented by $a$ in the formula $y^2=4ax$; so, in comparable notation, $(\star)$ becomes $|PQ|=4a\csc^2\theta$. $\square$