The equation of the line passing through origin which is parallel to the tangent of the curve $y=\dfrac{x-2}{x-3}$ at $x=4$ is

Question

The equation of the line passing through origin which is parallel to the tangent of the curve  $y=\dfrac{x-2}{x-3}$  at  $x=4$  is

cdquestions Admin · Accepted Answer

Given that: $y=\dfrac{x-2}{x-3}$ at  $x=4$ ,  $y=\dfrac{2}{1}=2$ Hence we can write, $\dfrac{dy}{dx}=\dfrac{(x-3)-(x-2)}{(x-3)^2}$ at  $ x=4$  ,   $\dfrac{dy}{dx}=-1$ Tangent can be represented as  $y-2=-1(x-4)$ $x + y − 6 = 0$ $x + y + k = 0$ Comparing both these equation we  found  $x+y=0$ $\therefore x=-y$  (_Ans)

Answer

$y=2x$

Answer

$y=2x+1$

Answer

$y=-x$

Answer

$y=x+2$

Answer

$y=4x$

The equation of the line passing through origin which is parallel to the tangent of the curve \(y=\dfrac{x-2}{x-3}\) at \(x=4\) is

The Correct Option is C

Solution and Explanation