JEE-Mains 2015 (12/30)

The equation for the curve can be written as

$x^{2} + 2xy + y^{2} - 4y^{2} = 0 \Longrightarrow (x + y)^{2} = (2y)^{2} \Longrightarrow x + y = \pm 2y.$

If $x + y = 2y$ then we have the line $y = x,$ and if $x + y = -2y$ then we have the line $y = -\frac{1}{3}x.$ So the given "curve" consists of two lines, one of slope $1$ and the other of slope $-\frac{1}{3},$ which intersect at the origin.

Now the point $(1,1)$ lies on the line $y = x,$ and so the normal to the curve at this point is a line of slope $-1$ with equation $y - 1 = -1*(x - 1) \Longrightarrow y = -x + 2.$ This normal will intersect the other component of the given curve, namely the line $y = -\frac{1}{3}x,$ when

$-x + 2 = -\dfrac{1}{3}x \Longrightarrow 2 = \dfrac{2}{3}x \Longrightarrow x = 3, y = -1.$

Thus the normal to the curve at $(1,1)$ meets the curve again at $(3,-1),$ i.e., in the fourth quadrant.