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- Thread starter suvadip
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- Jan 17, 2013

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You should start by looking at the points of intersections of lines ...

- Feb 29, 2012

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- Feb 7, 2012

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If the conic $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines then it must be of the form $(lx+my+n)(px+qy+r)=0$, where (comparing coefficients) $a = lp$, $h = \frac12(lq+mp)$, $b = mq$, ..., $c = nr.$ The conic $ax^2+2hxy+by^2-2gx-2fy+c=0$ is then given by $(lx+my-n)(px+qy-r)=0.$If each of the equations $ax^2+2hxy+by^2+2gx+2fy+c=0$ and $ax^2+2hxy+by^2-2gx-2fy+c=0$ represents a pair of straight lines, find the area of the parallelogram enclosed by them.

The answer to a question of this sort is almost sure to involve the discriminant $\Delta \mathrel{\overset{\text{def}}{=}}\begin{vmatrix}a&h \\h&b \end{vmatrix}$, so it would be worth finding this in terms of $l,m,n,p,q,r$. For a conic representing two straight lines, the discriminant is always negative, and in this case you can check that $$\Delta = \begin{vmatrix}lp & \tfrac12(lq+mp) \\ \tfrac12(lq+mp) & mq \end{vmatrix} = -\tfrac14(lq-mp)^2.$$ The line $lx+my+n = 0$ meets the lines $px+qy \pm r = 0$ at the points $\bigl(\frac{\pm mr-nq}{lq-mp},\frac{\pm lr-np}{lq-mp}\bigr)$, and the distance between those two points is $\left|\dfrac{2r\sqrt{l^2+m^2}}{lq-mp}\right|.$ So that is the length of one side of the parallelogram.

The perpendicular distance between the parallel lines $lx+my\pm n = 0$ is $\left|\dfrac{2n}{\sqrt{l^2+m^2}}\right|.$ So that is the distance between opposite sides of the parallelogram. The area $A$ of the parallelogram is therefore $$\left|\dfrac{2r\sqrt{l^2+m^2}}{lq-mp}\,\dfrac{2n}{\sqrt{l^2+m^2}}\right| = \left|\frac{2nr}{\frac12(lq-mp)}\right|.$$ In terms of the original coefficients, we can write this as $\boxed{A = \dfrac{2|c|}{\sqrt{-\Delta}}}.$

I think that there ought to be a more conceptual proof of this result, using only the original coefficients $a,h,b,g,f,c$, but I do not see one.

- Jan 29, 2012

- 1,151

I think you are misunderstanding the question. Each equation represents a

- Feb 29, 2012

- 342

Thank you Halls. I was!