(b) Let P(secθ,tan‌θ) be the point where the normal to the rectangular hyperbola x2−y2=1 Meet equation of normal at ‌
Ï€
4
is ‌‌
x
sec‌
Ï€
4
+‌
y
tan‌
Ï€
4
=2 ⇒‌
x
√2
+y=2 ⇒‌‌x+√2y=2√2 Substitute x=2√2−√2y in hyperbola's equation, ‌[√2(2−y)]2−y2=1 ⇒2[4+y2−4y]−y2=1 ⇒8+2y2−8y−y2−1=0 ⇒y2−8y+7=0 ⇒‌‌y=7,1 ‌x+7√2=2√2‌ or ‌x+√2=2√2 ⇒‌‌x=−5√2‌ or ‌x=√2 The two points are (−5√2,7),(√2,1). At (‌
Ï€
4
), the point is (√2,1) and at θ, the point is (−5√2,7). secθ=−5√2,tan‌θ=7 sec2θ+tan‌θ=(−5√2)2+7=57