Maths - The discriminant?

IF you are required to use the discriminant, here's a way:

Note first that the curve C: x^2+xy+2 = 0 is such that xy = -2 - x^2 < 0. So if x<0, then y>0, and if x>0, y<0.

In other words, C is comprised of two chunks, one in the 2nd quadrant, and the other in the 4th quadrant. This means the line y=x never intersects C (evident if you replace y by x in C !); but this means also that the line L: y = x + c only intersects one of the chunks, and never the other (for a given c).

Now, the line L: y = x+c is tangent to the curve C: x^2+xy+2 = 0 when it *at least* touches it. So replacing y by x+c in C:

*1) 0 = x^2 + x(x+c) + 2 = 2x^2 + cx + 2.

The discriminant of this quadratic is: c^2 - 16. But since we said L crosses C only ONCE for a given c, then we must have c^2-16 = 0, otherwise we would have two distinct roots for x in (*1), i.e. two values for which L intersects C.

So c^2=16, i.e. c=4 or c=-4.

(which would give the values x=-1 and x=1 respectively, and the lines y = x + 4 and y = x - 4 respectively)

(Of course, there's a gap in this 'proof': maybe C is so crooked that it meet the line L at more than two points, even in one of the quadrants. One can dismiss this possibility by analyzing C more in detail, e.g. by showing the line y = - x is an asymptote, and further that the derivative y ' of C is strictly increasing for x<0, and strictly decreasing for x>0.

But if we're allowed to differentiate, then one could solve the problem without using the discriminant:

we are looking for values x=a such that y ' (a) = 1.

Differentiating C: 0 = 2x + y + x y', and at x=a: 0 = 2a + y(a) + a*1 = y(a) + 3a, so y(a) = -3a.

So C has a derivative equal to 1 at (a, -3a).

Putting this point in C: 0 = a^2 - 3a^2 + 2 = - 2a^2 + 2, so a^2 = 1, i.e. a=1 and a= -1.

And (1, -3) and (-1, 3) are the points at which C has a tangent of slope equal to 1.

The corresponding lines are thus y = x - 4 and y = x + 4, which gives us c=4 and c= -4)