A) 16
B) 12
C) 8
D) 4
E) 2
Correct Answer: D
Solution :
Given curves are \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{12}=1\] \[\Rightarrow \] \[\frac{2x}{{{a}^{2}}}+\frac{2y}{12}.\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{12x}{{{a}^{2}}y}={{m}_{1}}\] (say) and \[{{y}^{3}}=8x\Rightarrow 3{{y}^{2}}\frac{dy}{dx}=8\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{8}{3{{y}^{2}}}={{m}_{2}}\] (say) For \[\theta =\frac{\pi }{2},1+{{m}_{1}}{{m}_{2}}=0\] \[\Rightarrow \] \[1+\left( \frac{-12x}{{{a}^{2}}y} \right)\left( \frac{8}{3{{y}^{2}}} \right)=0\] \[\Rightarrow \] \[3{{a}^{2}}(8x)-96x=0\] \[\Rightarrow \] \[{{a}^{2}}=4\]
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