Quadratic/Line Tangent Help

mwg83

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Mar 20, 2019
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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.

11441
 
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 \(\displaystyle x_1\).
 
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
 
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.
 
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