Question

find the equation of tangent to the curve Y equal to x minus 7 upon x minus 2 x minus 3 at the point where it cuts the x axis​

Answer 1

Answer:

Equation of tangent is x-20y-7=0

Step-by-step explanation:

Given curve is

$y=\dfrac{x-7}{(x-2)(x-3)}$

since it cuts the x axis, y=0

$\implies\,\dfrac{x-7}{(x-2)(x-3)}=0$

$\implies\,x-7=0$

$\implies\,x=7$

$\therefore\text{The given curve cuts the x axis at (7,0)}$

$y=\dfrac{x-7}{x^2-5x+6}$

Differtiate with respect to x

Applying quotient rule of diferentiation

$\boxed{\bf\dfrac{dy}{dx}=\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}}$

$\dfrac{dy}{dx}=\dfrac{(x^2-5x+6).1-(x-7)(2x-5)}{(x^2-5x+6)^2}$

$\text{Slope of tangent, m}$

$=(\dfrac{dy}{dx})_(7,0)$

$=\dfrac{(7^2-5(7)+6).1-(7-7)(2(7)-5)}{(7^2-5(7)+6)^2}$

$=\dfrac{(7^2-5(7)+6)}{(7^2-5(7)+6)^2}$

$=\dfrac{1}{7^2-5(7)+6}$

$=\dfrac{1}{49-35+6}$

$=\dfrac{1}{20}$

$\text{Slope of tangent, m}=\dfrac{1}{20}$

$\textbf{Equation of tangent is }$

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

$y-0=\dfrac{1}{20}(x-7)$

$y=\dfrac{1}{20}(x-7)$

$20y=(x-7)$

$\implies\boxed{\bf\,x-20y-7=0}$