Correct answer is (a) (-2, -2)
Easy explanation: Equation of the given hyperbola is, xy = 4 ……….(1)
Differentiating both side of (1) with respect to y, we get,
y*(dx/dy) + x(1) = 0
Or dx/dy = -(x/y)
Thus, the required equation of the normal to the hyperbola at (2, 2) is,
y – 2 = -[dx/dy](2, 2) (x – 2) = -(-2/2)(x – 2)
So, from here,
y – 2 = x – 2
Or x – y = 0 ……….(2)
Solving the equation (1) and (2) we get,
x = 2 and y = 2 or x = -2 and y = -2
Thus, the line (2) intersects the hyperbola (1) at (2, 2) and (-2, -2).
Hence, the evident is that the normal at (2, 2) to the hyperbola (1) again intersects it at (-2, -2).