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The result of a question in my book hinges on the fact that for $y^2 = 4ax$ every chord passing through (4a,0) subtends a right angle at the vertex. It is suppoesed to be a standard result, but I have never heard of it anywhere. Would someone please tell me if there is an easy way to get this result?

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  • $\begingroup$ Right angle with what? I didn't understand your question. Add a diagram to your question or some link please to make it clearer $\endgroup$ – Fareed Abi Farraj Apr 14 '19 at 12:53
  • $\begingroup$ The line joining origin to the end points of the chord passing through $(4a,0)$ are at right angles. $\endgroup$ – Dbchatto67 Apr 14 '19 at 12:56
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Let a chord of the parabola $y^2=4ax$ passing through the point $(4a,0)$ meets the parabola at the points $P(at^2,2at)$ and $Q(as^2,2as).$ Then the equation of the chord is $$\frac {y-2at} {x-at^2} = \frac {2a(t-s)} {a(t^2-s^2)}.$$ Since it passes through the point $(4a,0)$ we get $$\frac {0-2at} {4a-at^2} = \frac {2} {t+s}.$$ Simplifying we get $ts=-4.$ Now let the angle between $OP$ and $OQ$ be $\theta$ then we have $$\cot \theta = \frac {1+ \frac {2at} {at^2} \cdot \frac {2as} {as^2}} {\frac {2at} {at^2} - \frac {2as} {as^2}} = \frac {1 + \frac {4} {ts}} {2 \left (\frac 1 t - \frac 1 s \right ) } = \frac {ts+4} {2(s-t)}.$$ Now using the fact that $ts=-4$ we get $$\cot \theta = 0$$ since $t \neq s.$ This shows that $\theta = 90^{\circ}.$

This shows that any chord of the parabola $y^2=4ax$ passing through the point $(4a,0)$ subtends a right angle at the origin (note that the origin is the vertex of the parabola $y^2 = 4ax$).

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  • $\begingroup$ That's only one of the infinitely many chords passing through $(4a,0)$. $\endgroup$ – Aretino Apr 14 '19 at 17:13
  • $\begingroup$ @Aretino now check my edited answer whether it holds good or not. $\endgroup$ – Dbchatto67 Apr 14 '19 at 18:04
  • $\begingroup$ Computing $P\cdot Q$ and figuring out when that vanishes seems like it would be simpler than working with $\cot\theta$. $\endgroup$ – amd Apr 14 '19 at 19:41
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In the same vein as Dbchatto67’s answer, but doing the computations a bit differently: Take two arbitrary points $P=(at^2,2at)$ and $Q=(as^2,2as)$ on the parabola with $s\ne t$ and $s,t\ne0$. They are colinear with $(4a,0)$ iff $$\det\begin{bmatrix}at^2&2at&1\\4a&0&1\\as^2&2as&1\end{bmatrix} = 2a^2(s-t)(st+4)=0.$$ Since $a\ne0$, we must have $st+4=0$. On the other hand, since the vertex of the parabola is the origin, the segments joining it to these two points are orthogonal iff $$(at^2,2at)\cdot(as^2,2as) = a^2st(st+4)=0.$$ Instead of a dot product, one can work with distances: $OP^2+OQ^2=PQ^2$ iff $\angle{POQ}$ is a right angle. We have $$(at^2-as^2)^2+(2at-2as)^2-(a^2t^4+4a^2t^2)-(a^2s^4+4a^2s^2) = -2a^2st(st+4)=0.$$

Alternatively, there is a one-parameter family of lines through $(4a,0)$ that intersect the parabola in two points: $x=\lambda y+4a$. (We can take the coefficient of $x$ to be $1$ because the line $y=0$ intersects the parabola at only one point.) A straightforward application of the quadratic formula and a bit of elbow grease yields the intersection points $$x = 2\left(a(\lambda^2+2)\pm\lambda\sqrt{a^2(\lambda^2+4)}\right) \\ y = 2\left(a\lambda\pm\sqrt{a^2(\lambda^2+4)}\right).$$ As before, the segments joining these points to the origin are orthogonal iff the dot product of these two points vanishes: $$4a^2\left((\lambda^2+2)^2-\lambda^2(\lambda^2+4)\right)+4a^2\left(\lambda^2-(\lambda^2+4)\right) = 4a^2\left((\lambda^4+4\lambda^2+4)-(\lambda^4+4\lambda^2)-4\right) = 0.$$

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