Let S=3x2+2y2−5 The equation of pair of tangent drawn to the ellipse from the point (1,2) is, SS1=T2 Therefore, (3x2+2y2−5)(3(1)2+2(2)2−5)=(3x(1)+2y(2)−5)2 (3x2+2y2−5)6=(3x+4y−5)2 9x2−4y2−24xy+30x+40y−55=0 Hence, a=9 h=−12 b=−4 The angle between the line is calculated as, tanθ=