A) \[\frac{{{a}^{2}}{{x}_{1}}}{{{y}_{1}}}\]
B) \[\frac{{{a}^{2}}{{y}_{1}}}{{{x}_{1}}}\]
C) \[2{{a}^{2}}\]
D) \[4{{a}^{2}}\]
Correct Answer: C
Solution :
Since \[y=\frac{{{a}^{2}}}{x},\,\,\,\therefore \frac{dy}{dx}=-\frac{{{a}^{2}}}{{{x}^{2}}}\] \[\therefore \] At \[({{x}_{1}},\,{{y}_{1}}),\]\[\frac{dy}{dx}=\frac{-{{a}^{2}}}{x_{1}^{2}}\] Thus tangent to the curve will be \[y-{{y}_{1}}=\frac{-{{a}^{2}}}{x_{1}^{2}}(x-{{x}_{1}})\] Þ \[yx_{1}^{2}-{{y}_{1}}x_{1}^{2}=-{{a}^{2}}x+{{a}^{2}}{{x}_{1}}\] Þ \[y'\frac{1}{\sqrt{1-{{\left( \frac{2x}{1+{{x}^{2}}} \right)}^{2}}}}.\frac{2(1+{{x}^{2}})-4{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}}\], \[(\because {{x}_{1}}{{y}_{1}}={{a}^{2}})\] This meets the x-axis where \[y=0\] \[\therefore {{a}^{2}}x=2{{a}^{2}}{{x}_{1}},\]\[\therefore x=2{{x}_{1}}\] \[\therefore \]Point on the x-axis is \[(2{{x}_{1}},0)\] Again tangent meets the y-axis where \[x=0\] \[\because x_{1}^{2}y=2{{a}^{2}}{{x}_{1}},\,\ \ \ \therefore y=\frac{2{{a}^{2}}}{{{x}_{1}}}\] So point on the y-axis is \[\left( 0,\frac{2{{a}^{2}}}{{{x}_{1}}} \right)\] Required area \[=\frac{1}{2}(2{{x}_{1}})\left( {{\frac{2a}{{{x}_{1}}}}^{2}} \right)=2{{a}^{2}}\].
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