Hyperbola

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)

2⋅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=‌

−1

2

So, β=4α−30
‌=4(‌

−1

2

)−30=−32
‌∴|β|+‌

1

|α|

=|−32|+‌

1

|−1∕2|

‌‌‌=32+2=34