Explanation: Given curves, y2=4x and xy=c cuts orthogonally.
Let they intersect at (x1,y1).
Now,
∴‌‌2y‌

dy

dx

=4
⇒‌‌‌

dy

dx

=‌

2

y

⇒

dy

dx

|(x1,y1)=‌

2

y1

‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
‌ and ‌‌‌xy=c
∴‌‌x‌

dy

dx

+y=0
‌∴‌‌

dy

dx

‌=‌

−y

x

⇒‌

dy

dx

|(x1,y1)‌=−‌

y1

x1

‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
From eqs. (i) and (ii)
‌‌

2

y1

×(‌

−y1

x1

)‌=−1‌‌[∵m1m2=−1]
⇒‌x1‌=2
‌ Put ‌‌x1‌=2‌ in ‌y12=4x1‌, we get ‌
⇒y12‌=4(2)=8
‌y1‌=2√2
Now, put value of x1 and y1 in x1y1=c, we get
c=2(2√2)=4√2