A) Straight line
B) Parabola
C) Ellipse
D) Circle
Correct Answer: D
Solution :
Variable line is\[\frac{x}{a}+\frac{y}{b}=1\] ... (i) Any line perpendicular to (i) and passing through the origin will be \[\frac{x}{b}-\frac{y}{a}=0\] ... (ii) Now foot of the perpendicular from the origin to line (i) is the point of intersection (i) and (ii) Let it be\[P(\alpha ,\,\,\beta )\],then \[\frac{\alpha }{a}+\frac{\beta }{b}=1\] ... (iii) and \[\frac{\alpha }{b}-\frac{\beta }{a}=1\] ? (iv) Squaring and adding Eqs. (iii) and (iv), we get \[{{\alpha }^{2}}\left( \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}} \right)+{{\beta }^{2}}\left( \frac{1}{{{b}^{2}}}+\frac{1}{{{a}^{2}}} \right)=1\] Hence, the locus of\[P(\alpha ,\,\,\beta )\]is\[{{x}^{2}}+{{y}^{2}}={{c}^{2}}\] which is a circle.
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