Given hyperbola equation is xy=16 Differentiate it w.r.t x, we get ‌y+x⋅‌
dy
dx
‌=0 ⇒‌‌
dy
dx
‌=‌
−y
x
Now, slope of tangent at (8,2) is m1=‌
−2
8
=‌
−1
4
And slope of normal is mn=‌
−1
−1∕4
=4 Now, equation of normal at (8,2) is ‌y−2=4(x−8) ⇒‌‌y=4x−30 Normal intersects the hyperbola again at a point (α,β), So, αβ=16 and β=4α−30 Solving these two equations, we get ‌α(4α−30)=16 ‌⇒‌‌2α2−15α−8=0 So,α‌=‌
−(−15)±√(−15)2−4⋅2⋅(−8)
2â‹…2
‌=‌
15±√289
4
=‌
15±17
4
So, α1=‌
15+17
4
=‌
32
4
=8 and α2=‌
15−17
4
=‌
−1
2
Since (8,2) is one point, the other point is where α=‌
