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  • State whether the statements are true or false. The line x/a + y/b= 1moves in such a way that 1/a^{2}+1/b^{2}=1/c^{2} , where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x^{2 }+ y^{2} = c^{2}

State whether the statements are true or false.

The line \frac{x}{a}+\frac{y}{b}=1 moves in such a way that \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}, where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x^{2}+y^{2}=c^{2}

Answers (1)

Equation of line \frac{x}{a}+\frac{y}{b}=1……i   

 Equation of line passing through the origin and perpendicular to the given line \frac{x}{a}-\frac{y}{b}=0….ii

 Now the foot of perpendicular from origin on the line (i) is the point of intersection of lines (i) and ii 

So, to find its locus we have to eliminate the variable a and b  

 Squaring and adding both the equations we get\left (\frac{x}{a}+\frac{y}{b} \right )^{2}+\left (\frac{x}{a}-\frac{y}{b} \right )^{2}=1+0

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{2xy}{ab}+\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{2xy}{ab}=1

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

x^{2}\left (\frac{1}{a^{2}}+\frac{1}{b^{2}} \right )+y^{2}\left (\frac{1}{a^{2}}+\frac{1}{b^{2}} \right )=1  

  \left (x^{2}+y^{2} \right )+\left (\frac{1}{a^{2}}+\frac{1}{b^{2}} \right )=1

\left (x^{2}+y^{2} \right )+\left (\frac{1}{c^{2}} \right )=1

  x2+y2=c2  

Hence, the given statement is true.

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