The number of tangents to the curve xy-3y+2=0 having slope 2 is:

The correct answer is Option (3) → 2

The number of such tangents is the number of points on the curve where dy/dx = 2. From the equation of the curve,

    \(y(x - 3) = -2\)

    \( \displaystyle y = -\frac{2}{x - 3} \)

    \( \displaystyle \frac{dy}{dx} = \frac{2}{(x - 3)^2} \)

Setting the slope equal to 2:

    \( \displaystyle \frac{2}{(x - 3)^2} = 2 \)

    \((x - 3)^2 = 1,\quad x = 4 \text{ or } x = 2\)

The points are (4, −2) and (2, 2). The tangents there are \(y = 2x - 10\) and \(y = 2x - 2\), two different lines. So there are 2 tangents with slope 2.