Question

A rod of length $l$ sides with its ends on two perpendicular lines. Fins the locus of its mid points.

• A. $x^2+y^2=l^2$
• B. $x^2+y^2=\frac{l^2}{4}$
• C. $2x^2+2y^2=l^2$
• D. $x^2+y^2=2l^2$

Answer 1

Answer: (A)

$x^2+y^2=l^2$

$x=2\alpha$

$y=2\beta$

$x^2+y^2=l^2$

$\Rightarrow (2\alpha {)}^2+(2\beta {)}^2=l^2$

$=4{\alpha }^2+4{\beta }^2=l^2$

${\alpha }^2+{\beta }^2=\frac{l^2}{4}$

$\Rightarrow \frac{x^2}{4}+\frac{y^2}{4}=\frac{l^2}{4}$

$\Rightarrow x^2+y^2=l^2$