The given bends are \[y\text{ }=\text{ }4\text{ }\text{ }x2\text{ }\ldots \text{ }.\text{ }\left( I \right)\text{ }and\text{ }y\text{ }=\text{ }x2\text{ }\ldots \text{ }\left( ii \right)\]
Also, we realize that the point of convergence of two bends is equivalent to the point between the digressions attracted to the bends at their place of convergence.
Presently, separating conditions (I) and (ii) w.r.t x, we have
\[dy/dx\text{ }=\text{ }-\text{ }2x\Rightarrow m1\text{ }=\text{ }-\text{ }2x\]
m1 is the incline of the digression to the bend (I).
also, \[dy/dx\text{ }=\text{ }2x\Rightarrow m2\text{ }=\text{ }2x\]
m2 is the incline of the digression to the bend (ii).
Thus, \[m1\text{ }=\text{ }-\text{ }2x\text{ }and\text{ }m2\text{ }=\text{ }2x\]
On tackling condition (I) and (ii), we get
\[4\text{ }\text{ }x2\text{ }=\text{ }x2\Rightarrow 2×2\text{ }=\text{ }4\Rightarrow x2\text{ }=\text{ }2\Rightarrow x\text{ }=\text{ }\pm \surd 2\]
Thus, \[m1\text{ }=\text{ }-\text{ }2x\text{ }=\text{ }-\text{ }2\surd 2\text{ }and\text{ }m2\text{ }=\text{ }2x\text{ }=\text{ }2\surd 2\]
Let θ be the point of convergence of two bends