Quadratic/Line Tangent Help

[mwg83]

I'm trying to connect a quadratic ax^2+bx and a line cx+d with an arc of radius r

CAD programs do it so simply but to come up with an equation for the arc has me lost. I basically need to find the circle center location that lies on a normal line of the quadratic that intersects a line offset to the main line by distance r.

 

[HallsofIvy]

Yes, that is exactly what you want to do. Given the line $\displaystyle y= cx+ d$ (don't forget the "y") a line perpendicular to it at $\displaystyle x= x_0$ is of the form $\displaystyle y= (-1/c)(x- x_0)+ cx_0+ d$. Given the parabola $\displaystyle y= ax^2+ bx$ a line perpendicular to it at $\displaystyle x= x_1$ is of the form $\displaystyle y= (-1/(2x_1+ b))(x- x_1)+ ax_1^2+ bx_1$. To determine the point where the two lines intersect, solve $\displaystyle (-1/c)(x- x_0)+ cx_0+ d=(-1/(2x_1+ b))(x- x_1)+ ax_1^2+ bx_1$ for x. Determine the distances from $\displaystyle x_0$ and $\displaystyle x_1$ to the point of intersection and set them equal to determine $\displaystyle x_0$ and $\displaystyle x_1$.

 

[mwg83]

Thank you for the reply, I'm still a little confused. Are you just saying plot everything out and see where they intersect? What are the values of x0 and x1

 

[HallsofIvy]

I didn't say anything about plotting! I told you how to find $\displaystyle x_1$ and $\displaystyle x_0$. So you want the answer, not how to find it yourself? That's tedious but pretty straight forward. I said "solve $\displaystyle (−1/c)(x−x_0)+cx_0+d=(−1/(2x_1+b))(x−x_1)+ax^2_1+bx_1$". That's a linear equation to be solved for x. Do you know how to solve a linear equation.