Question

Find the equation of tangent to the curve $y=\frac{x-7}{(x-3)(x-4)}$ at the point where it cut the x -axis.

Answer 1

Given $y=\frac{x-7}{(x-3)(x-4)}$

The point where it cut x-axis will be $(x,0)$, y-coordinate will be zero

$\frac{(x-7)}{(x-3)(x-4)}=0\Rightarrow x-7=0$

$x=7$

$y=\frac{(x-7)}{(x-3)(x-4)}=(x-7)[\frac{1}{x-4}-\frac{1}{x-3}]=\frac{x-7}{x-4}-\frac{x-7}{x-3}$

$\frac{dy}{dx}=\frac{(x-4)-(x-7)}{(x-4{)}^2}-\frac{(x-3)-(x-7)}{(x-3{)}^2}=\frac{3}{(x-4{)}^2}-\frac{4}{(x-3{)}^2}$

Slope $\frac{dy}{dx}$ at $x=7$ is $\frac{3}{(7-4{)}^2}-\frac{4}{(7-3{)}^2}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$

The tangent drawn at $(7,0)$ and having slop $\frac{1}{12}$

$(y-y_1)=m(x-x_1)$

$(y-0)=\frac{1}{12}(x-7)$

$x-12y=7$

$\therefore$ The equation of a tangent to the curve at the point where it cut x-axis is $x-12y=7$