Tangent Line Problem--Please Help

**doctorgk** · November 10th 2007, 06:12 PM

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.

**ThePerfectHacker** · November 10th 2007, 06:20 PM

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

**doctorgk** · November 11th 2007, 03:27 AM

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

**kalagota** · November 11th 2007, 04:54 AM

Originally Posted by ThePerfectHacker

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

the other solution to the system is (-2,-1), not (-2,1)

**ThePerfectHacker** · November 11th 2007, 07:59 AM

Originally Posted by doctorgk

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

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

the other solution to the system is (-2,-1), not (-2,1)

Okay I fix it.

**doctorgk** · November 11th 2007, 02:54 PM

makes sense now, thank you very much

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Tangent Line Problem--Please Help

Okay but still a bit lost

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