Question

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

Answer 1

Given, equation of curve is

$y=\frac{(x-7)}{(x-2)(x-3)}$ ....(1)

To find the intersection of given curve with X-axis, Put $y=0$ in equation 1, we get,

$0=\frac{x-7}{(x-2)(x-3)}$

$x-7=0⟹x=7$

Thus, the curve cut the X-axis at (7,0).

Now, on differentiating equation of curve w.r.t. x, we get,

$\frac{dy}{dx}=\frac{(x-2)(x-3).1-(x-7)\left[\right(x-2).1+(x-3\left).1\right]}{\left[\right(x-2\left)\right(x-3){]}^2}$

$=\frac{(x-2)(x-3)-(x-7)(2x-5)}{\left[\right(x-2\left)\right(x-3){]}^2}$

$=\frac{(x-2)(x-3)[1-\frac{(x-7)}{(x-2)(x-3)}(2x-5\left)\right]}{\left[\right(x-2\left)\right(x-3){]}^2}$

$=\frac{1-y(2x-5)}{(x-2)(x-3)}$

$(\frac{dy}{dx}{)}_{(7,0)}=\frac{1}{20}$

Thus, slope of tangent = $\frac{1}{20}$

Slope of normal = $-20$

Hence, the equation of tangent at (7,0) is

$y-0=\frac{1}{20}(x-7)⟹20y-x+7=0$

And, the equation of normal at $(7,0)$ is

$y-0=-20(x-7)⟹20x+y-140=0$