A) \[a\]
B) \[2a\]
C) \[2\sqrt{a}\]
D) None of these
Correct Answer: A
Solution :
\[\sqrt{x}+\sqrt{y}=a\]; \[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0\], \ \[\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}\] Hence tangent at (x, y) is \[Y-y=-\frac{\sqrt{y}}{\sqrt{x}}(X-x)\] or \[X\sqrt{y}+Y\sqrt{x}=\sqrt{xy}\,\,\left( \sqrt{x}+\sqrt{y} \right)=\sqrt{axy}\] or \[\frac{X}{\sqrt{a}\sqrt{x}}+\frac{Y}{\sqrt{a}\sqrt{y}}=1\]. Clearly its intercepts on the axes are \[\sqrt{a}\sqrt{x}\] and \[\sqrt{a}\sqrt{y}\]. Sum of the intercepts = \[\sqrt{a}\left( \sqrt{x}+\sqrt{y} \right)=\sqrt{a}.\sqrt{a}=a\].
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