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Problem:If $t_1$ and $t_2$ are roots of the equation $t^2+kt+1=0$ , where $k$ is an arbitrary constant. Then prove that the line joining the points $(at_1^2,2at_1),(at_2^2,2at_2)$ always passes through a fixed point.Also find that point.

I have done this question by a fairly difficult procedure(at least I think so) ..by finding the equation of the line passing through the given points as $$y(t_2+t_1)-2at_1t_2-2at_1^2=2x-2at_1^2$$ And then using the relation from the quadratic equation given and writing it as $$(x+a)+k(\frac{y}{2})=0$$ which is the equation of family of lines and so which gives the fixed point as $(-a,0)$.

But actually I found this question in exercises of parabola and so I was thinking if it was possible to solve this question without this family of lines and all that (that is by using geometrical properties of parabola)...I have been thinking about this ...and I could only notice that since $(at^2,2at)$ is the parametric equation of a general point on a parabola $y^2=4ax$ so the points $(at_1^2,2at_1),(at_2^2,2at_2)$ can be thought of as two given points on parabola $y^2=4ax$ where $t_1t_2=1$ but I am not able to get that fixed point (which apparently is a point on the directrix of parabola $y^2=4ax$ from the above method that is $(-a,0)$) based on the above information by using the geometrical properties of parabola. Any help is appreciated on this.

Freelancer
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2 Answers2

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enter image description here

For real roots, $|k| \ge 2$. Also, note that the slope of the line through the two points is given by, $m=\frac{2}{t_1+t_2}=-\frac{2}{k}$, which therefore lies in the range $|m|\le1 $. For the various value of $k$ in the specified range, this is a family of lines which connect two points of a parabola (recall the parametric representation of a parabola) whose focus is at $(a,0)$ and latus rectum is $4a$. The lines are tangent in the case $k=\pm2$, i.e, $m=\mp1$ since in these cases there is only one root of the quadratic. Then it is sufficient to find the point of intersection of these tangents, which is easily found to be $(-a,0)$. For, any other line that passes through this point and has a slope $m$ such that $ 0 \lt |m|\lt 1 $ clearly meets the parabola at two points. That such lines can be parametrized by $|k| \ge 2$ is verified by showing that $(-a,0)$, $(at_1^2,2at_1)$ and $(\frac{a}{t_1^2}, \frac{2a}{t_1})$ are colinear.

vnd
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  • How did you get the condition $|k|>=2$ ? Can you please explain how you got this relation.? – Freelancer Feb 24 '16 at 12:35
  • discriminant of the quadratic should be non-negative for real roots. – vnd Feb 24 '16 at 12:37
  • "the lines are tangent in the case $k=+-2$"....I don't understand which lines are you talking about..is it the family of lines,or the tangents made at the points $(at_1^2,2at_1),(at_2^2,2at_2)$ can you please add a rough Image for representing the lines that you ate talking about.... – Freelancer Feb 25 '16 at 03:29
  • http://i.stack.imgur.com/mO4ey.jpg – Freelancer Feb 25 '16 at 03:38
  • I have added the above image for a rough idea ..but I am still unable to get which lines are you talking about. – Freelancer Feb 25 '16 at 03:40
  • For example, when $k=-2$, the quadratic becomes $t^2-2t+1=0$. Both the roots of this equation are same, i.e, $t_1=t_2=1$. In other words, the line with slope $m=\frac{2}{t_1+t_2}=1$ will touch the parabola at the point $(at_1^2,2at_1)\equiv (a,2a)$ tangentially. It is this line that I was referring to, the other line being the corresponding to the case $k=2$. – vnd Feb 25 '16 at 04:29
  • So this is actually something like this http://i.stack.imgur.com/HmXo0.jpg – Freelancer Feb 25 '16 at 06:16
  • yes. exactly that. – vnd Feb 25 '16 at 06:17
  • Thanks ...I have got the answer ...But just as a sidenote have we actually proved in this question that if we make a line joining any two points on the prarbola such that its slope is <=1 then that line will always pas through the point $(-a,0)$ ? – Freelancer Feb 25 '16 at 06:21
  • Let us continue this discussion in chat. – vnd Feb 25 '16 at 06:22
  • @Freelancer The x- coordinate of the point of intersection at left is $ a t_1t_2$. I have given above $t_1t_2 = -1$ with correct sign so always the tangents cut on the directrix. – Narasimham Feb 26 '16 at 08:32
  • @Narasimham I very well appreciate the point you make, but the problem statement nowhere indicates that the chord through $(at_1^2,2at_1),(at_2^2,2at_2)$ passes through focus. – vnd Feb 26 '16 at 09:19
  • @ vnd Yes, indeed it does.. determine and definitively fix or locate "a" fixed point to be "the" focus fixed point and its invariant position in between the given ends of focal chord $(at_1^2,2at_1),(at_2^2,2at_2)$ and whose $t_1t_2$ product has to be $-1$, The single center point on directrix is an incorrect invariant to be a fixed point. – Narasimham Feb 26 '16 at 09:59
  • @Narasimham But the condition $t_1t_2=-1$ would ensure what you are saying. Here the condition is $t_1t_2=1$. If you go through the above solution, $(-a,0)$ satisfies the criterion of the problem. – vnd Feb 26 '16 at 10:08
  • @ vnd Yes, agreed. I concede to this (to me) a new fixed point. A single graphic showing both pencil of rays would be great. – Narasimham Feb 26 '16 at 15:23
  • @ vnd I am still wondering why both did n't get included automatically in a differential search formulation for fixed point. – Narasimham Feb 26 '16 at 15:58
  • @Narasimham Please see the uploaded image. For the differential search, you are differentiating the equation $(x- a)+k(\frac{y}{2})=0$, wrt $k$ to begin with. – vnd Feb 26 '16 at 18:05
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Tgts at ends Parb

With some reference to the above:

When the roots are solved, the tangent drawn at these particular slope pairs all cut on the directrix when the chord between $(t_1- t_2)$ points of tangency passes through the focus.

EDIT 1

There appears to be an incorrect tangent/slope supplying equation to start with, at the outset:

$$ t^2 + k t + 1 = 0 $$

whereas the correct equation for the given parametrization ought to be

$$ t^2 + k t - 1 = 0 $$

EDIT 3:

( It is now conceded the former sign is correct for all rays through (-a,0) )

By change of sign in the constant term we have two different "foci"

The sum of roots at either end of a parabola can be arbitrary, but since tangent at either end of focal chord are perpendicular ( meeting on directrix but that is besides the point) their product of slopes must be necessarily $-1$ and not $+1$ as we can see directly from quadratic equation coefficients.

This property of parabola is easy to verify and shall not do it here.

Taking correct sign means chaging sign of $ a, a\rightarrow -a $

$$ (x- a)+k(\frac{y}{2})=0 $$

EDIT 2:

( To find fixed point by C discriminant , partially differentiate w.r.t. k, y=0, and so (x=a, y=0) is the fixed point).

which is the equation of all focal rays passing through a fixed point, viz., the focus of parabola.

(BTW, It will be also an interesting problem to see how the (red) directrix line is fixed)

I am giving a sketch for confirmatory reference.

ParabolaTgts&FixedPtFocusenter image description here

Narasimham
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