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

• A. $y=2x$
• B. $y=2x+1$
• C. $y=-x$
• D. $y=x+2$
• E. $y=4x$

Answer 1

Answer: (C)

$y=-x$

Answer 2

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)