question_answer

Let a and b be any two non-zero real numbers satisfying \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{4}\]. Then,  the  foot  of perpendicular drawn from origin on the variable line \[\frac{x}{a}+\frac{y}{b}=1\], lies on

A)  circle of radius 2.

B)  parabola of length of latus-rectum 4.

C)  ellipse with length of semi-major axis 2.   

D)  hyperbola with length of semi-transverse axis \[\sqrt{2}\].

Correct Answer: A

Solution :

\[\frac{x}{a}+\frac{y}{b}=1\]               ?(1) \[\therefore \] Equation of line is \[hx+ky={{h}^{2}}+{{k}^{2}}\] So, on comparing (1) and (2), we get \[\frac{1}{a}\,=\frac{h}{{{h}^{2}}+{{k}^{2}}}\,;\,\frac{1}{b}\,=\frac{k}{{{h}^{2}}+{{k}^{2}}}\] As, \[\frac{1}{{{a}^{2}}}\,+\frac{1}{{{b}^{2}}}\,=\frac{1}{4}\] \[\Rightarrow \,\,{{h}^{2}}+{{k}^{2}}=4,\] which is circle of radius 2.