A) 1
B) -1
C) 2
D) 3
Correct Answer: D
Solution :
Given equation of curves are \[xy=2\] ?(i) and \[{{x}^{2}}+4y=0\] ?(ii) On solving Eqs. (i) and (ii), we get point of intersection is \[-(2,-1).\] From Eq. (i) \[x\frac{dy}{dx}+y=0\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{y}{x}\] \[{{m}_{1}}={{\left( \frac{dy}{dx} \right)}_{(-2,-1)}}=-\frac{1}{2}\] and from Eq. (ii), \[2x+4\frac{dy}{dx}=0\] \[\Rightarrow \] \[{{m}_{2}}=1\] \[\because \] \[\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{-\frac{1}{2}-1}{1-\frac{1}{2}} \right|=\left| \frac{\frac{3}{2}}{\frac{1}{2}} \right|\] \[\therefore \] \[\tan \theta =3\]
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