question_answer

Let the   tangent   to the \[\sqrt{x}+\sqrt{y}=\sqrt{a}\,\,(a>0)\] curve at any point on it cuts the coordinate axes at P and Q. Then \[OP\text{ }+\text{ }OQ,\] where 0 is origin is equal to-

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 \[(0,{{y}_{1}}+\sqrt{{{x}_{1}}{{y}_{1}}},)\]

\[\therefore OP+OQ={{x}_{1}}+{{y}_{1}}+2\sqrt{{{x}_{1}}{{y}_{1}}}\]

\[={{(\sqrt{{{x}_{1}}}+\sqrt{{{y}_{1}}})}^{2}}=a\]