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Let ax-y+7=0 be a chord of parabola x^2=28y meeting it at A and B ,and tangents at A and B meet at C.the locus of circumcentre of tr.ABC is
Hello student,Given chord is ax – y + 7 = 0 and the parabola is x2 = 28y.We first try to find out the points of intersection of the chord with the parabola as it would give us the end points of the chord.So, substituting the value of y = ax+7 from the equation of the chord into the equation of the parabola we have,x2 = 28(ax+7)So, x = 14(a ± √a2+1)y = x2/28Let us denote the x and y cordinates as x*x** and y*y**.Then the slope of the tangents at these two points would be given bym*m** which would be equal to the value of dy/dx at x*x**.Hence, m*m** = x*x**/14.This clearly shows that the product of slopes is -1.Hence, this means that the two tangents are perpendicular and therefore the triangle is a right angled triangle.Circumcentre of a right angled traingle is given by the mid-point of the hypotenuse.So, the coordinates of circumcentre are x$ = 14ay$ = (x*2 + x**2)/ 28 = 7(2a2+1)
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