# Algebra, points of intersection

• Feb 16th 2013, 06:40 PM
elmidge
Algebra, points of intersection
Why does 'a' have 2 values if y= x is tangent to parabola y = x^2 + ax + 1. I thought there would be only 1 solution.

Similarly, why are there 2 valus of 'm' when y = mx - 0.5 is tangent to y = x^2 + 5x?

There is only 1 solution for b when y = -x is tangent to y = x^2 + x + b ??
• Feb 16th 2013, 09:45 PM
ibdutt
Re: Algebra, points of intersection
Please explain how did you proceed?
• Feb 17th 2013, 12:59 AM
elmidge
Re: Algebra, points of intersection
I used the discriminant to find the unknown. I equated the discriminant to 0 as the line is a tangent to the parabola.
• Feb 17th 2013, 05:04 AM
HallsofIvy
Re: Algebra, points of intersection
In order that a line, y= mx+ n, be tangent to a parabola, \$\displaystyle y= ax^2+ bx+ c\$, the line must first intersect the parabola: the point, (x, y), must satisfy \$\displaystyle mx+n= ax^2+ bx+ c\$. But in order that it be tangent, that must be a double root which means that the discriminant of the equation \$\displaystyle ax^2+ (b- m)x+ c- n= 0\$ must be 0. The discriminant, \$\displaystyle (b- m)^2- 4a(c- n)\$ is quadratic in both b and m and so that equation may have two roots, while c, in \$\displaystyle y= x^2+ x+ c\$ is not quadratic.

Geometrically, increasing a, in the parabola, \$\displaystyle y= x^2+ ax+ 1\$, moves the vertex to the left and downward. There will be two different values of a, one positive, the other negative, that will be tangent to y= x.
I have attached a graph.
Attachment 27084

For the second problem, y= mx+ 0.5 can be any line passing through (0, 0.5). That point is NOT on the given parabola so there exist two different lines through it that are tangent to the parabola, one above y= 0 and one below y= 0.
Attachment 27085

Here is a graph of the last one:
Attachment 27086