Assume the point of intersection be (x0,y0) x2−y2=4 Differentiate the above equation with respect to x. 2x−2y
dy
dx
=0
dy
dx
==
x
y
At point (x0,y0) m1=
x0
y0
and curve x2+y2=4√2 Differentiate with respect to x 2x+2y
dy
dx
=0
dy
dx
=
x
y
At point (x0,y0) m2=−
x0
y0
The angle between the tangents is calculated as, θ=tan−1|
x0
y0
+
x0
y0
1+
x0
y0
(−
x0
y0
)
| =tan−1(
x0y0
2
) since, the point (x0,y0) lies on curves then, x02−y02=4 and x02+y02=4√2 Then, x0y0=2 …… (2) Substitute the value in equation (1), θ=tan−1(