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 the 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)

$x^2+5y^2+4xy-1=0$

Let the point on the positive x axis be $(a,0)$, on the line $x+y=0$ be $(-b,b)$ and midpoint be $(h,k)$
Co-ordinates of $h$ will be $\frac{a-b}{2}$ and that of k will be $\frac{b}{2}$ [as it is midpoint of AB ]
AB is of fixed length 2 units $\therefore (a+b{)}^2+b^2=4$
Thus $a^2+2ab+2b^2=4$ $\to a^2-2ab+b^2+4ab+b^2=4$
$(a-b{)}^2+4ab+b^2=4$ Substituting $a-b=2h$ and $b=2k$ and $a=2k+2h$ we get
$(2h{)}^2+4(2k+2h\left)\right(2k)+(2k{)}^2=4$ which simplifies to $h^2+5k^2+4hk-1=0$
Therefore, locus of midpoint is $x^2+5y^2+4xy-1=0$
Hence, option 'A' is correct.