A) a
B) 2a
C) \[{{a}^{2}}\]
D) \[\sqrt{a}\]
Correct Answer: A
Solution :
\[\because \sqrt{x}+\sqrt{y}=\sqrt{a}\] |
Slope of tangent at \[({{x}_{1}},{{y}_{1}})\] is |
\[\frac{1}{2}\frac{1}{\sqrt{{{x}_{1}}}}+\frac{1}{2}\frac{1}{\sqrt{{{y}_{1}}}}\left( \frac{dy}{dx} \right)=0\] |
\[\Rightarrow \frac{dy}{dx}=\frac{-\sqrt{{{y}_{1}}}}{\sqrt{{{x}_{1}}}}\] |
Equation of tangent is \[y-{{y}_{1}}=\frac{-\sqrt{{{y}_{1}}}}{\sqrt{{{x}_{1}}}}(x-{{x}_{1}})\] |
Point P on x-axis is \[({{x}_{1}}+\sqrt{{{x}_{1}}{{y}_{1}}},0)\] |
Point Q on y-axis is \[(0,{{y}_{1}}+\sqrt{{{x}_{1}}{{y}_{1}}},)\] |
\[\therefore OP+OQ={{x}_{1}}+{{y}_{1}}+2\sqrt{{{x}_{1}}{{y}_{1}}}\] |
\[={{(\sqrt{{{x}_{1}}}+\sqrt{{{y}_{1}}})}^{2}}=a\] |
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