question_answer

Let a and b be any two numbers satisfying\[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{4}.\] Then, the foot of perpendicular from the origin on the variable line,\[\frac{x}{a}+\frac{y}{b}=1,\]lies on:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

A) a hyperbola with each semi-axis \[=\sqrt{2}\]

B) a hyperbola with each semi-axis = 2

C) a circle of radius = 2

D) a circle of radius \[=\sqrt{2}\]

Correct Answer: C

Solution :

Let the foot of the perpendicular from (0, 0) on the variable line \[\frac{x}{a}+\frac{y}{b}=1\] is \[({{x}_{1}}>{{y}_{1}})\] Hence, perpendicular distance of the variable line\[\frac{x}{a}+\frac{y}{b}=1\] from the point O (0, 0) = OA \[\Rightarrow \]\[\frac{|-1|}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}}=\sqrt{x_{1}^{2}+y_{1}^{2}}\] \[\Rightarrow \]\[\frac{1}{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}=x_{1}^{2}+y_{1}^{2}\] \[\Rightarrow \]\[4=x_{1}^{2}+y_{1}^{2}\left[ \because \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{4} \right],\] which is equation of a circle with radius 2. Hence \[({{x}_{1}},{{y}_{1}})\]i.e., the foot of the perpendicular from the point (0, 0) to the variable line \[\frac{x}{a}+\frac{x}{b}=1\]is lies on a circle with radius = 2