3
$\begingroup$

Suppose I have any second degree conic and a point $P$ lying within the conic. A family of lines pass through this point $P$. We have to find the line for which the area of the corresponding section is minimised. And separately, we want to find a second line, for which the perimeter is minimised. Given is a example traced out in desmos, where we want to minimise the area $[ABC]$ and separately minimise perimeter of triangle $ABC$.

enter image description here

And here is another example, to illustrate that we want to minimize the area and perimeter for a general second degree conic, showing the situation for a parabola. Where we want to find the inclination of the straight line through P so that the perimeter of the yellow part is maximum. And we want to find a separate condition so that the area is maximum (These are two questions). enter image description here

I decided to work on a simpler problem first, that of a pair of straight lines. One may quite easily workout the case by using the general equation of a pair of straight lines, that is given by

$$ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$

and solving for points in common with $y = mx + c$, and just minimising the determinant, perimeter may be worked out using the distance formula. However a similar method doesn't seem plausible for the other cases. I thought a geometrical interpetration of the same might be a bit better.

My question is, how can we interpret and solve this problem by geometrical or visual arguments? If not, how can we generalise the same for a general second degree conic?

The problem was posed in a high school examination, and is thought to have a nice geometrical answer, which I haven't been able to figure out.

EDIT: After fiddling a bit, I figured out that for a pair of straight lines, the minimum perimeter is achieved when the line touches a circle which has the pair of tangents given by the pair of straight lines, I suppose considering a non Euclidean plane would help to handle the other cases geometrically? My attempts at doing it have failed.

$\endgroup$
4
  • $\begingroup$ It is not clear to me which area you want to minimize. $\endgroup$ Nov 12, 2023 at 11:15
  • $\begingroup$ @Intelligentipauca I have added more information and another example $\endgroup$ Nov 13, 2023 at 11:44
  • $\begingroup$ Just intuitively/by symmetry, it would seem that to minimize area, $AB$ should be the line parallel to the directrix for a parabola, or either the major or minor axis which passes through $P$. $\endgroup$ Nov 13, 2023 at 11:52
  • $\begingroup$ The problem is for a general conic, although I suppose such intuitive methods are the way to go. I do not however understand how intuition would lead the way for perimeter. $\endgroup$ Nov 13, 2023 at 12:10

2 Answers 2

1
$\begingroup$

Parabola case.

Without loss of generality we can assume that the parabola is $y=ax^2,$ $a>0$. Let $P=(x_0,y_0)$. Let the line passing through $P$ has slope $m$ and cuts the parabola at $-x_1<0$ and $x_2>0$. Then the area of the bounded region between the line and the parabola is $$A=a\frac{x_1^2+x_2^2}2(x_2+x_1)-a\frac{x_2^3+x_1^3}3=a\frac{(x_2+x_1)^3}6.$$ So we need to minimize $x_2+x_1=\frac1a\sqrt\Delta$ where $$\Delta=m^2-4ax_0m+4ay_0.$$ So $m=2ax_0$.

$\endgroup$
1
+100
$\begingroup$

I have the following heuristic arguments for the area case.

For the simplicity assume that the point $P$ is the origin $(0,0)$. Suppose that the line $\ell$ passing through the point $P$ and points $P_1$ and $P_2$ on the conic, provides the local extremum of the cut area. Then $|PP_1|=|PP_2|$ otherwise we can both increase and decrease the cut area, rotating $\ell$ around $P$ on a small angle. Thus $P=\frac{P_1+P_2}2$ and so $P_1=-P_2$. Let $P_1=(x_1,y_1)$. Plugging the coordinates of $P_1$ and $P_2$ into the equation of the conic we obtain that $gx_1+fy_1=0$ and $ax_1^2 + 2hx_1y_1 + by^2_1 + c = 0$. In the nondegenerated case this system should lead to a quadratic equation for $x_1$ or $y_1$.

More technical approach is to affinely transform the conic to one of canonical forms. Since the areas are changed proportionally, it suffices to consider canonical forms.

Parabola. It was already considered in Bob Dobbs' answer.

Ellipse. The canonical conic is the circle and the line crossing $P$ cuts a segment from the circle. Since the area of the cut segment is monotonically increasing function of the length $|P_1P_2|$ of its chord $P_1P_2$, we have to minimize this chord length. Since the product $|PP_1|\cdot |PP_2|$ is constant, by the inequality between the arithmetic and geometric means, the length $|P_1P_2|=|P_1P|+|PP_2|$ attains maximum when $|P_1P|=|P_2P|$.

Hyperbola. Choosing the respective coordinate system, we can ensure that the equation of the hyperbola is $y=\frac 1x$, $P=(x_0,y_0)$, and $x_0,y_0>0$. Suppose that a line $\ell$ passes through $P$ and points $P_1=\left(x_1,\frac 1{x_1}\right)$ and $P_1=\left(x_2,\frac 1{x_2}\right)$ of the hyperbola such that $x_1<x_2$. Then the cut area equals $$\frac{1/x_1+1/x_2}{2}(x_2-x_1)-\int_{x_1}^{x_2}\frac{dx}{x}=\frac{x_2^2-x_1^2}{x_1 x_2}-\ln\frac{x_2}{x_1}.$$ ...

$\endgroup$
4
  • $\begingroup$ I have awarded the bounty to this answer, even though it isnt a complete answer to my problem (it is complete for the area case, however the perimeter case is more puzzling), since it would be awarded to the other answer automatically I suppose $\endgroup$ Nov 21, 2023 at 10:07
  • $\begingroup$ @ShivanshJaiswal As far as I remember, the bounty is awarded automatically either to an accepted answer or, if there is no accepted answer, then, I guess, the half-size bounty is awarded to the most scored answer, provided it has at least +2 score. From the other hand, my opinion concerning bounty awards is that in order to confirm that the bounty offer was fair and to acknowledge efforts of all bounty aspirants, a bounty should be given to the best answer, provided it is more or less advanced, although it can be imperfect or incomplete. $\endgroup$ Nov 22, 2023 at 5:16
  • $\begingroup$ Finally, it seems that the perimeter case can be considered similarly to the area case, but we need to formulate the condition of a local extremum for it. $\endgroup$ Nov 22, 2023 at 5:16
  • $\begingroup$ Please check my edit. $\endgroup$ Nov 22, 2023 at 19:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .