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.
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Same doubt. The first answer's method is easily available online but I want to know a method using geometry. – Clemens Bartholdy Jul 14 '21 at 06:09
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here – Clemens Bartholdy Jul 14 '21 at 06:09
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The answer is $4a\csc^2 \alpha$, where $\alpha$ is angle of chord with x axis – Clemens Bartholdy Jul 14 '21 at 06:11
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1You are right, I thought about it and my edit invalidates the first answer to the question. When I first changed the title, I thought it would be harmless since the asker left the site. @Blue – Clemens Bartholdy Jul 14 '21 at 06:24
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1@Buraian: FYI, I have posted a geometric solution. – Blue Jul 14 '21 at 09:00
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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 $|PQ|=4a\csc^2\theta$. $\square$
Blue
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How is the horizontal pink line equal to $2|FF'| \csc \theta$, rest I understand – Clemens Bartholdy Aug 05 '21 at 08:55
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@Buraian: As mentioned, $\square FPP'M$ and $\square FQQ'M$ are right angled kites, which imply $\overline{MF}\cong\overline{MP'}$ and $\overline{MF}\cong\overline{MQ'}$, hence $|P'Q'|=2|MF|$. If it's not clear why the quadrilaterals are kites, draw $\overline{MP}$ and $\overline{MQ}$ to break them into pairs of right angles; each pair has a congruent legs and a common hypotenuse, so are congruent triangles. That $|MF|=|FF'|\csc\theta$ is elementary right-triangle trigonometry in $\triangle MFF'$. ... Does that help? – Blue Aug 05 '21 at 23:51
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Ok I read it again with a fresh mind, firstly I don't get why two side pair equal + angle equal implies that last two sides are equal in the quadrilateral (and hence the claim it is right angled kite) – Clemens Bartholdy Aug 06 '21 at 08:45
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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?
John
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