Question

The normal to the curve $y(x-2)(x-3)=x+6$ at the point where the curve intersects the $y$-axis passes through the point

[1 mark]

• A. $(-\frac{1}{2},-\frac{1}{2})$
• B. $(\frac{1}{2},\frac{1}{2})$
• C. $(\frac{1}{2},-\frac{1}{3})$
• D. $(\frac{1}{2},\frac{1}{3})$

Answer 1

The correct option is B $(\frac{1}{2},\frac{1}{2})$

$y=\frac{x+6}{(x-2)(x-3)}$
Coordinates of point of intersection with $y$-axis is $(0,1)$
$y=\frac{x+6}{x^2-5x+6}$
$\Rightarrow y^{\prime }=\frac{(x^2-5x+6)-(2x-5)(x+6)}{(x^2-5x+6{)}^2}$
$\Rightarrow y^{\prime }{|}_{x=0}=\frac{6-(-30)}{36}=1$
Then, slope of normal at $(0,1)$ is $-1$
Equation of normal passing through $(0,1)$ is $y-1=-1(x-0)$
$\text{i.e.},x+y=1$
Thus, the normal passes through $(\frac{1}{2},\frac{1}{2})$.