Question

A line of fixed length $2$ units moves so that its ends are on the positive $x$-axis and that part of the line $x+y=0$ which lies in the second quadrant. Then the locus of the midpoint of the line has equation

• A. $x^2+5y^2+4xy-1=0$
• B. $x^2+5y^2+4xy+1=0$
• C. $x^2+5y^2-4xy-1=0$
• D. $4x^2+5y^2+4xy+1=0$

Answer 1

Answer: (A)

If $\angle BAO=\theta$, then $BM=2sin\theta$
and $MO=BM=2sin\theta$,
$MA=2cos\theta$. Hence
$A=(2cos\theta -2sin\theta ,0)$
and $B=(-2sin\theta ,2sin\theta )$.
Since $P(x,y)$ is the midpoint of $AB$,
we have $2x=(2cos\theta )+(-4sin\theta )$
or $cos\theta -2sin\theta =x$
$2y=(2sin\theta )$ or $sin\theta =y$
Eliminating $\theta$, we have
$(x+2y{)}^2+y^2=1$ or $x^2+5y^2+4xy-1=0$