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Q.
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
Application of Derivatives
Solution:
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
$\frac{d y}{d x} =\frac{(x-2)(x-3)(1)-(x-7)(2 x-5)}{(x-2)^2(x-3)^2}$
$\frac{d y}{d x} =\frac{1-\frac{(x-7)(2 x-5)}{(x-2)(x-3)}}{(x-2)(x-3)} $
$\frac{d y}{d x} =\frac{1-y(2 x-5)}{(x-2)(x-3)}$
or $\left. \frac{d y}{d x}\right]_{(7,0)} =\frac{1-0}{(5)(4)}=\frac{1}{20}$
Therefore, the slope of the tangent at $(7,0)$ is $\frac{1}{20}$.
Hence, the equation of the tangent at $(7,0)$ is
$y-0=\frac{1}{20}(x-7) \text { or } 20 y-x+7=0$