Question

A rod of length $2$ units moves so that one end is on x-axis and other end on $y+x=0$. Then find the equation of locus of the mid-point of the rod.

Answer 1

$\to$ As one and rod lines on $x$ axis, hence coordinate be $A(x,0)$

Let other end be $B(x_2,y_2)$

$(x_2,y_2)$ line on $x+y=0$ then $x_2+y_2=0$

Therefore $y_2=-x_2$

Hence point $B(x_2-x_2)$

Let mid point of $AB$ be $(m,n)$

thus $m=\frac{x_1+x_2}{2}$

$2x=x_1+x_2$ equation $\left(1\right)$

now $n=\frac{-x_2+0}{2}$

$2n=-x_2$

$-2n=x_2$ equation $\left(2\right)$

Put equation $\left(2\right)$ in equation $\left(1\right)$

$2m=x_1-2n$

$x_1=2m+2n$

Given that, $AB=2$

$A{B}^2=4$

$(x_1-x_2)+(0-(-x_2){)}^2=4$

$(x_1-x_2{)}^2+x_2=4$

$(2m+2n+2n{)}^2+(-2n{)}^2=4$

$(2m+4n{)}^2+4n^2=4$

$4(m+2n{)}^2+4n^2=4$ equation ((3)$

$(m+2n{)}^2+n^2=1$ (Divide equation $\left(3\right)$ by $4$)

$m^2+4n^2+4mn+n^2=1$

$m^2+4mn+5n^2=1$

Replace $m$ by $x$ and $n$ by $y$

Therefore $x^2+4xy+5y^2=1$