tangents drawn to the parabola y^2=8x at the points p(t1) and q(t2) intersect at a point R such that t1 and t2 are the roots of the equation t^2+at+2=0, |a|>2root2. find locus of circumcenter of triangle PTQ is ans y square =x-2 Share with your friends Share 20 Manbar Singh answered this The given equation of the parabola is,y2 = 8x ........1The above equation is of the form of y2 = 4ax.So, 4a = 8 ⇒a = 2Now the parametric coordinates of any point on 1 are at2,2at or 2t2,4t.Now, coordinates of the point P are P2t12, 4t1 and Q are P2t22, 4t2.Now, differentiating 1, 2ydydx = 8⇒dydx = 4yNow, slope of the tangent at point P = 44t1 = 1t1slope of the tangent at point Q = 44t2 = 1t2Now, equation of RP is, y - 4t1x - 2t12 = 1t1⇒t1y - 4t12 = x - 2t12⇒t1y - x = 2t12 ......2Now, the equation of RQ is, t2y - x = 2t22 ........3Solving 2 and 3, x = 2t1t2; y = 2t1+t2Now, we have, t2 + at + 2 = 0Since t1 and t2 are the roots of above equation, sot1 + t2 = -at1t2 = 2So, x = 4; y = -2aSo, the coordinates of the point R are : 4, -2a.Now, circumcentre of ∆PQR is 4+2t12+2t222, -2a+4t1+4t22≡a2-2, -3a -47 View Full Answer . answered this plsssssssd anssssssssss... its urgent 6