• Nov 10th 2007, 06:12 PM
doctorgk
Here is how the problem reads, the graphing part I got, it's just the problem itself.

Graph the two parabolas y=x^2 and y=-x^2 +2x -5 in the same coordinate plane. Find the equations of the two lines simultaneously tangent to both parabolas.

I got as far as just finding the derivatives which as 2x and -2x+2, but I can't the equations.
• Nov 10th 2007, 06:20 PM
ThePerfectHacker
Quote:

Originally Posted by doctorgk
Here is how the problem reads, the graphing part I got, it's just the problem itself.

Graph the two parabolas y=x^2 and y=-x^2 +2x -5 in the same coordinate plane. Find the equations of the two lines simultaneously tangent to both parabolas.

I got as far as just finding the derivatives which as 2x and -2x+2, but I can't the equations.

Given an parabola \$\displaystyle y=ax^2+bx+c,a\not = 0\$ and a line \$\displaystyle y=px+q\$. This line is tangent to the parabola if and only if \$\displaystyle (b-p)^2 - 4a(c-q)=0\$. So if \$\displaystyle px+q\$ is tangent to \$\displaystyle x^2\$ then it means \$\displaystyle p^2 + 4q = 0\$ and it tangent to \$\displaystyle -x^2+2x-5\$ that means \$\displaystyle (p-2)^2 - 4(q+5)=0\$. The solutions to this system of equations is \$\displaystyle (p,q) = \{(-2,-1),(4,-4)\}\$.
• Nov 11th 2007, 03:27 AM
doctorgk
Okay but still a bit lost
how did you get that formula? is there a way to solve this with derivatives?
• Nov 11th 2007, 04:54 AM
kalagota
Quote:

Originally Posted by ThePerfectHacker
Given an parabola \$\displaystyle y=ax^2+bx+c,a\not = 0\$ and a line \$\displaystyle y=px+q\$. This line is tangent to the parabola if and only if \$\displaystyle (b-p)^2 - 4a(c-q)=0\$. So if \$\displaystyle px+q\$ is tangent to \$\displaystyle x^2\$ then it means \$\displaystyle p^2 + 4q = 0\$ and it tangent to \$\displaystyle -x^2+2x-5\$ that means \$\displaystyle (p-2)^2 - 4(q+5)=0\$. The solutions to this system of equations is \$\displaystyle (p,q) = \{(-2,1),(4,-4)\}\$.

the other solution to the system is (-2,-1), not (-2,1)
• Nov 11th 2007, 07:59 AM
ThePerfectHacker
Quote:

Originally Posted by doctorgk
how did you get that formula? is there a way to solve this with derivatives?

Given a parabola \$\displaystyle y=ax^2+bx+c,a\not = 0\$ and \$\displaystyle y=px+q\$ if they intersect exactly once then the equation (of the intersection) \$\displaystyle ax^2+bx+c = px+q\$ has exactly one solution, this is a quadradic so, \$\displaystyle ax^2+(b-p)x+(c-q)=0\$. We want this quadradic to has exactly one real solution that happens when the discrimant is zero.

(We want it to have exactly one solution because a tangent line to a parabola intersects it exactly one time. )

Quote:

the other solution to the system is (-2,-1), not (-2,1)
Okay I fix it.
• Nov 11th 2007, 02:54 PM
doctorgk
Okay
makes sense now, thank you very much