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 If the line \mathrm{x / a+y / b=1} moves in such a way that \mathrm{1 / a^2+1 / b^2=1 / c^2}, where c is a constant, then the foot of perpendicular from the origin on the straight line describes the circle:

Option: 1

\mathrm{x^2+y^2=c^2}


Option: 2

\mathrm{x^2+y^2=2 c^2}


Option: 3

\mathrm{x^2+y^2=4 c^2}


Option: 4

\mathrm{2\left(x^2+y^2\right)=c^2}


Answers (1)

best_answer

The given line is  \mathrm{\text { } x / a+y / b=1\ \ \ \ ............\left ( 1 \right )}
Where \mathrm{1 / \mathrm{a}^2+1 / \mathrm{b}^2=1 / \mathrm{c}^2\ \ \ \ .......\left ( 2 \right )}
Equation of the line perpendicular to \mathrm{\left ( 1 \right )} and passing through origin is \mathrm{x / b-y / a=0\ \ \ \ \ .........\left ( 3 \right )}

The locus of the foot of perpendicular from the origin on the straight line \mathrm{\left ( 1 \right )}, i. e. the point of intersection of \mathrm{\left ( 1 \right )} and \mathrm{\left ( 3 \right )}, is obtained by eliminating a and b between \mathrm{\left ( 1 \right )} and \mathrm{\left ( 3 \right )}, using \mathrm{\left ( 2 \right )}.
For this squaring and adding \mathrm{\left ( 1 \right )} and \mathrm{\left ( 3 \right )}, we have \mathrm{\left(1 / a^2+1 / b^2\right) x^2+\left(1 / a^2+1 / b^2\right) y^2=1}
or \mathrm{\frac{\mathrm{x}^2}{\mathrm{c}^2}+\frac{\mathrm{y}^2}{\mathrm{c}^2}=1,\ \ \ \ \quad[\operatorname{using}(2)]}
or \mathrm{x^2+y^2=c^2} which is the required locus.

 

Posted by

Sanket Gandhi

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