Math Help Forum - View Single Post

Soroban · #2 June 23rd, 2009, 05:09 AM

Hello, usagi_killer!

Quote:

Find the values of $p$ such that $y=px$ intersects $y=x^2 + 1$ twice.

Explain the geometric significance of this.

The parabola and line intersect: . $x^2+1 \:=\:px \quad\Rightarrow\quad x^2 - px + 1 \:=\:0$

. . Quadratic Formula: . $x \:=\:\frac{p \pm \sqrt{p^2-4}}{2}$

With two intersections, the discriminant must be positive:
. . $p^2 - 4 \:>\:0 \quad\Rightarrow\quad p^2 \:>\:4 \quad\Rightarrow\quad |p| \:>\:2$

Hence: . $p \:<\:-2\;\text{ or }\;p \:>\:2$

Make a sketch.
We have an up-opening parabola with vertex (0,1)
. . and a straight line through the origin.

Code:

        *       |       *
                |
         *      |      *
          *     |     *
            *   |   *     o
               1*       o
                |     o
                |   o
                | o
      - - - - - + - - - - - - - 
              o |
            o   |
                |

If the slope $p$ is +2 or -2, the line is tangent to the parabola.
. . There is one intersection.

If $p$ is greater than -2 and less than 2, the line misses the parabola.
. . There is no intersection.

If $p$ is less than -2 or greater than 2, there are two intersections.