Hyperbola

Since Normal at point P makes equal intercept on co-ordinate axes, therefore slope of Normal =−1
Hence slope of tangent =1
Equation of tangent
y−0=1(x−1)
y=x−1
Equation of tangent at (x1y1)
‌

xx1

a2

−‌

yy1

b2

=1
x−y=1 (equation of Tangent)
on comparing x1=a2,y1−b2
Also a2−b2=1‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)
Now equation of normal at (x1y1)=(a12b2)
y−b2=−1(x−a2)
x+y=a2+b2... (Normal)
point of intersection with x -axis is (a2+b2)
Now e=√1+‌

b2

a2

e=√1+‌

b2

b2+1

[ from (1)‌

b2

b2+1

<1]
1<e<√2 option (A)

∆=‌

1

2

⋅AB⋅PQ
∆=‌

1

2

⋅AB⋅PQ
‌ and ‌∆=‌

1

2

(a2+b2−1)⋅b2
∆=‌

1

2

(2b2)b2‌ (from ‌‌ (1) ‌a2−1=b2‌ ) ‌
∆=b4 so option ( D )