Question: Below is the question attachment In an effort to improve angioplasty, in which a wire is threaded through a coronary artery to guide a balloon

Below is the question attachment

Below is the question attachment In an effort to improve angioplasty, in

In an effort to improve angioplasty, in which a wire is threaded through a coronary artery to guide a balloon which then opens a blocked artery, researchers have modelled the guidewire using a quadratic equations y = px- + qx+r Part of the process involves taking pairs of points (Xo, yo) and (X1, )1 ) on a X-Ray image and finding the values of p,q and r described below: [a.] Substitute the two points in the quadratic equation, and call the resulting equations Eo and E1 . Subtract E1 - Eo to eliminate r and solve the resultant equation to find p. p = 1 - yo - q(x1 - XO) [b.] Substitute the result from part (a) into the equation Eo and solve for r to obtain r = yo - XO y1 - yo Xi - XO + qx 7 X1 - XO X3 - XO - qxo [c.] Substitute the values of p and r from part (a) and (c) into the original quadratic equation to get y V1 - yo x + q x - xo X1 - XO + 1 - XO X - XO XO + yo - y1 - yo x3 - XO XO [d.] Substitute K = y1 - yo x7 - X- L = Kx- and M =. X1 - XO X1 + Xo into result of part (c) to obtain y = Kx + q(x - Xo - Mx2 + Mx3)+ yo-L [e.] Finally, substitute A = L - yo P = Mxo , and R = X0 - P, into the result from part (d), and solve for q to obtain q = y - Kx2 + A x - R - Mx2

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