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

The correct answer is Option (3) → 2

Given curve

$xy-3y+2=0$

Differentiate implicitly w.r.t. $x$

$x\frac{dy}{dx}+y-3\frac{dy}{dx}=0$

Group terms

$(x-3)\frac{dy}{dx}+y=0$

$\frac{dy}{dx}=\frac{-y}{x-3}$

Given slope $=2$

$2=\frac{-y}{x-3}$

$y=-2(x-3)$

$y=-2x+6$

This line must touch the curve, so substitute in curve equation

$x(-2x+6)-3(-2x+6)+2=0$

$-2x^2+6x+6x-18+2=0$

$-2x^2+12x-16=0$

$x^2-6x+8=0$

Discriminant

$D=36-32=4>0$

Two real and distinct points of contact exist.

Number of tangents $=2$