y=x2−1 ⇒y=8x−x2−9 For point of intersection, x2−1=8x−x2−9 2x2−8x+8=0 ⇒x2−4x+4=0. ⇒(x−2)2=0 ⇒x=2 ∴y=22−1=3Point of intersection =(2,3) [
dy
dx
]c1=2x and [
dy
dx
]c2=8−2x At(2,3) [
dy
dx
]c1=4 and [
dy
dx
]c2=4
∴tan‌θ=|
[
dy
dxc1
]−[
dy
dx
]c2
1+[
dy
dx
]c1
[
dy
dx
]c2|...|θ→Angle‌between‌both‌curves| =0 Hence, both curves tough each other at (2,3)