# Find the equation of all lines having slope $$2$$ that are tangents to the curve $y= \large\frac{1}{x-3}, $$\: x \neq 3 ## 1 Answer Toolbox: • If y=f(x),then \big(\large\frac{dy}{dx}\big)_P=slope of the tangent to y=f(x) at the point P. • Equation of line with given two points is \large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1} Step 1: Given : y=\large\frac{1}{x-3} Differentiating w.r.t x we get, \large\frac{dy}{dx}=\large\frac{-1}{(x-3)^2}----(1) It is given the slope of the tangent is 2. Hence equating \large\frac{dy}{dx} to 2 we get, \large\frac{-1}{(x-3)^2}=$$2$
$\Rightarrow -1=2(x-1)^2$
Step 2:
On expanding we get,
$2x^2-12x+18=-1$
$\Rightarrow 2x^2-12+19=0$
Finding a solution for $x$ by using the quadratic formula $\large\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$\large\frac{12\pm\sqrt{144-4\times 2\times 19}}{4}=\large\frac{12\pm\sqrt{144-158}}{4}$
It has no possible solution.
So the curve has no tangent.