Question

The equation of the tangent to the curve $y=\frac{x-7}{(x-2)(x-3)}$ at the point, where it cuts the $x$-axis, is

• A. $20y-x+7=0$
• B. $20y+x-7=0$
• C. $10y-x+14=0$
• D. $15y+x-14=0$

Answer 1

Answer: (A)

Note that on X-axis, $y=0$. So, the equation of the curve, when $y=0$, gives $x=7$. Thus, the curve cuts the $x$-axis at $(7,0)$. Now, differentiating the equation of the curve with respect to $x$, we obtain
$\frac{dy}{dx}=\frac{(x-2)(x-3)(1)-(x-7)(2x-5)}{(x-2{)}^2(x-3{)}^2}$
$\frac{dy}{dx}=\frac{1-\frac{(x-7)(2x-5)}{(x-2)(x-3)}}{(x-2)(x-3)}$
$\frac{dy}{dx}=\frac{1-y(2x-5)}{(x-2)(x-3)}$
or ${\frac{dy}{dx}]}_{(7,0)}=\frac{1-0}{(5)(4)}=\frac{1}{20}$
Therefore, the slope of the tangent at $(7,0)$ is $\frac{1}{20}$.
Hence, the equation of the tangent at $(7,0)$ is
$y-0=\frac{1}{20}(x-7)\text{ or }20y-x+7=0$