Find the equation of the tangent and normal to the curve $y =\frac{x-7}{(x-2)(x-3)}$ at the point where it cuts the x-axis.

Tangent: $x−20y−7=0$; Normal: $20x+y−140=0$

The correct answer is Option (1) → Tangent: $x−20y−7=0$; Normal: $20x+y−140=0$

The given curve is $y =\frac{x-7}{(x-2)(x-3)}$   ...(i)

It cuts the x-axis i.e. $y = 0$ when $\frac{x-7}{(x-2)(x-3)}=0⇒x=7$.

Thus, the curve cuts the x-axis at the point $P(7, 0)$

Diff. (i) w.r.t. x, we get

$\frac{dy}{dx}=\frac{(x-2) (x-3).1-(x-7) (2x-5)}{((x-2)(x-3))^2}$.

∴ The slope of tangent at $P(7, 0) =\frac{5.4-0.9}{(5.4)^2}=\frac{1}{20}$.

∴ The equation of the tangent to the given curve at P(7, 0) is

$y-0=\frac{1}{20}(x-7)$ i.e. $x-20y-7=0$.

Slope of normal to the given curve at P(7, 0) is -20.

∴ The equation of the normal to the given curve at P(7, 0) is

$y-0=-20(x-7)$ i.e. $20x + y - 140 = 0$.