Question

let from (1,1) tangents are drawn to x^2-3y^2=1 then find the length of intercept made by these tangents on x+y=1

Answer 1

The equations of the tangents and normal at a point P(x1, y1) on the ellipse x2/a2+ y2/b2 = 1 arexx1/a2 + yy1/b2 = 1                                    ...... (1)and (x-x1)/((x1/a2)) + (y-y2)/((y1/b2)) = 1       ...... (2)Also the minor axis is y-axis i.e. x = 0Solving (1) and x = 0, we have A (0, b2/y1)Solving (2) and x = 0, we have B (0, y1 - a2y1/b2)Let S(ae, 0) be one of the foci of the ellipse. Then the slope of SA = ((b2/y1) - 0)/(0-ae)= b2/aey1 = m1 (say)And the slope of SB = ([y1 (a2/b2 ) y1 ]-0)/(0-ae)   = y1/ae ((a2-b2 ))/b2= (y1 a2 e2)/(aeb2) [·.· b2 = a2(1 - e2)]   = aey1/b2 = m2  (say)Evidently m1m2 = -1 ⇒ SA | SB