The line y + 2x = p has no intersection with the parabola y

^{2}= x + p. Prove that p <

**-**1/24.

Whenever there is an intersection, the two y co-ordinates must be equal, i.e., y

_{L}= y_{P}. According to the usual method, Mary lets y = p – 2x, so y^{2}= (p – 2x)^{2}= p^{2}– 4px +4x^{2}.y

^{2 }= y^{2}, so p^{2}– 4px +4x^{2}= x + pThen, 4x

^{2}– 4px + p^{2}– x – p = 04x

^{2}– (4p + 1) x + (p^{2}– p) = 0Finally, Mary uses the discriminant, b

^{2 }– 4ac, with a =4, b = – (4p + 1) and c = p^{2}– p.b

^{2 }– 4ac < 0(– (4p + 1))

^{2}– 4(4)(p^{2}– p) < 016p

^{2 }+ 8p + 1 – 16p^{2}+ 16p <024p < -1

p < -1/24

However, it must also be true that, whenever there is an intersection, the two x co-ordinates must be equal, i.e., x

_{L}= x_{P}. Therefore Johnny writes:Johnny also has a quadratic equation, and he also uses the discriminant. However, he has somewhat ‘nicer’ values for a, b and c. a = 1, b =½ and c = -1.5p.

b

^{2 }– 4ac < 0(½)

^{2}– 4(1)(-1.5p) < 0¼ + 6p <0

6p < -¼

p < -1/24

If you believe that every second counts on the exam, it pays to notice that y

^{2}= x + p is not our typical parabola, as it opens rightwards (instead of upwards or downwards).This observation ought to give rise to the notion that working with x

_{L}= x_{P }is likely to be easier than working with y_{L}= y_{P}.