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

Answer 1

Answer: (A)

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