Parallelogram area

[onta...@hotmail.com]

Find the two values of f for which the equation
8x^2 + 6xy - 5y^2 + 4x + 2fy - 4 = 0 represents straight lines. Find the area of the parallelogram formed from the four lines so obtained.

[João Pedro Afonso]

Lets try a blind guess (meaning I did the math, but didn't double check the results later... that's so wrong):

f is -9/2 or 6

and the area is 1.28571...

Nice problem,
JPA

[onta...@hotmail.com]

Right. What were the four linear equations ?
Bill

[João Pedro Afonso]

>Right. What were the four linear equations ?

y=-1 + 2x
y= 2 + 2x
y=-0.8 - 0.8 x
y= 0.4 - 0.8 x

The original equation is quadratic in y, so we use the proper formula to solve it. The resultant expression is linear in x plus or minus a square root of a polynomial (the discriminant) of degree 2. The only thing we need to do is to turn that polynomial in a square of another polynomial (In Portugal, we called it a remarkable case, we study 3 in school), which leads to another quadratic equation in f... and two solutions from one times the the solutions from the other, gives the 4 pretended linear equations. The rest is history (finding 3 intercession points, get 2 vectors from them, do a cross product to find the area).

[onta...@hotmail.com]

Yes I agree. I had the y's in terms of x's but after swiching around they are the same as yours.
Cheers
Bill

[onta...@hotmail.com]

Correction, should be x's in terms of the y's.

[hei...@duizendknoop.com]

I prefer an area of 18/sqrt(205) in stead of an unendless series of
little points.

[João Pedro Afonso]

> Correction, should be x's in terms of the y's.

Why you would do that? The usual convention is to put the Y's in terms of X's.

hmmm... forget it, if I put the X's in terms of Y's, the discriminant is going to appear linear in f, instead of quadratic... much more easy the first step that way. I didn't remember to do things that way (and was surprised when you said I was right... easy to do mistakes the way I did).

[onta...@hotmail.com]

9/7 acceptable ?

[João Pedro Afonso]

But yours is an exact result, which takes the opportunity from others to come and present a result with better accuracy. Although yours is very different from mine (But Bill's appears to be exactly mine... which means he did keep the numbers in fractional form until the end... I'm feeling ashamed of myself :-)