Find the coordinates of a point P where all the variable chords of the curve $y^2=8x$ subtending right angles at the origin are concurrent

Question

Let the chord be QR

$$Q(2t_1^2,4t_1)$$ and $$R(2t_2^2,4t_2)$$

When chord sub tends right angle at the origin, $$t_1t_2=-4$$

Also the equation of the chord is $$y-4t_1=\frac{4(t_2-t_1)}{2(t_2^2-t_1^2)}(x-2t_1^2)$$

When finding the common point of intersection, the equation is written in the form of $L_1+\lambda L_2$,

$$t_1^2y-2t_1x-4y+16t_1=0$$

What will the common point be in this case?

What you are looking for is not a single point P but a line or curve which is the locus of all possible points P. — sammy gerbil, Apr 05 '20 at 14:42
Sorry I misunderstood the question. ... P will lie on the axis of symmetry of the parabola. — sammy gerbil, Apr 05 '20 at 14:50

Quanto · Accepted Answer · 2020-04-05T16:03:29.230

1

The concurrent, or the fixed point $P$, can be identified from the vertical chord whose endpoints are symmetric with respect to the $x$-axis. Knowing that $t_1t_2 = -4$, we have $t_1=-t_2=2$. The fixed point is then $(2\cdot 2^2,0)=(8,0)$.

It remains to show that $(8,0)$ satisfies the chord equation,

$$\frac{y-4t_1}{x-2t_1^2}=\frac{4(t_2-t_1)}{2(t_2^2-t_1^2)}$$

that is, $$\frac{0-4t_1}{8-2t_1^2}=\frac2{t_1+t_2}\implies \frac{-t_1}{4-t_1^2 } = \frac1{t_1+t_2}$$

which is true due to $t_1t_2 = -4$.

edited Apr 05 '20 at 16:03

answered Apr 05 '20 at 15:34

Quanto

97,352

Why would they be symmetric? – Aditya Apr 05 '20 at 17:08
@Aditya - because the chord is vertical, perpendicular to the x-axis – Quanto Apr 05 '20 at 17:09
It subtends a right angle at the center. I understand that a vertical chord also satisfies that condition, but it is a specialized case, isn’t it? – Aditya Apr 05 '20 at 17:22
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@Aditya - yes, it is special chord to help locate the concurrent point – Quanto Apr 05 '20 at 17:27

Find the coordinates of a point P where all the variable chords of the curve $y^2=8x$ subtending right angles at the origin are concurrent

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