Let a and b be positive real numbers such that a>1 and b

Since Normal at point P makes equal intercept on co-ordinate axes, therefore slope of Normal = –1Hence slope of tangent = 1Equation of tangenty−0=1(x−1)y=x−1 Equation of tangent at x1y1xx1a2−yy1b2=1x−y=1 (equation of Tangent)  on comparing x1=a2,y1=b2 Also a 2−b2=1---(1) Now equation of normal at x1y1=a21b2y−b2=−1x−a2x+y=a2+b2… (Normal)  point of intersection with x -axis is a2+b2 Now e=1+b2a2e=1+b2b2+1from⁡(1)b2b2+1<11