A) \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]
B) \[y={{e}^{-x/2a}}\]
C) \[y=ax\]
D) \[{{x}^{2}}=4ay\]
Correct Answer: B
Solution :
\[{{y}^{2}}=4ax\]Þ\[2y{{\left( \frac{dy}{dx} \right)}_{1}}=4a\]Þ \[{{\left( \frac{dy}{dx} \right)}_{1}}=\frac{2a}{y}\] ?..(i) Taking curve \[y={{e}^{-x/2a}}\] \[{{\left( \frac{dy}{dx} \right)}_{2}}={{e}^{-x/2a}}\left( -\frac{1}{2a} \right)\]\[=-\frac{y}{2a}\] .....(ii) Both curves cut orthogonally if, \[{{\left( \frac{dy}{dx} \right)}_{1}}{{\left( \frac{dy}{dx} \right)}_{2}}=-1\]Þ\[\left( -\frac{y}{2a} \right).\left( \frac{2a}{y} \right)=-1\].
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