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 :

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

Answer 1

Answer: (A)

$y(x-2)(x-3)=x+6$
At y-axis, $x=0,y=1$
Now, on differentiation.
$\frac{dy}{dx}\left(x-2\right)\left(x-3\right)+y\left(2x-5\right)=1$
$\frac{dy}{dx}\left(6\right)+1\left(-5\right)=1$
$\frac{dy}{dx}=\frac{6}{6}=1$
Now slope of normal $=-1$
Equation of normal $y-1=-(x-0)$
$y+x-1=0$... (i)
Line(i) passes through $\left(\frac{1}{2},\frac{1}{2}\right)$