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

The correct option is A

$\left(\frac{1}{2},\frac{1}{2}\right)$

Explanation for the correct answer

Given curve, $y(x-2)(x-3)=x+6$

Substitute $x=0$ to find intersection of curve with $y$ axis.

$\Rightarrow$$y\left(-2\right)\left(-3\right)=6$

$\Rightarrow$ $y=1$

$\Rightarrow$The intersection of the curve with $y$ axis is at $x=0$,$y=1$

Differentiating with respect to $x$

$\frac{dy}{dx}\left(x-2\right)\left(x-3\right)+y\left(2x-5\right)=1$

Substitute $x=0$,$y=1$ to obtain the value of slope at the intersection of the curve and $y$ axis

$\frac{dy}{dx}\left(-2\right)\left(-3\right)+1\left(-5\right)=1$

$\Rightarrow$ $\frac{dy}{dx}=1$

Tangent is perpendicular to the normal.

Hence the slope of the normal is $-1$.

The equation for the normal is given by using the slope point form of a line

$y-1=-1(x-0)$

$x+y-1=0$

From the given options $\left(\frac{1}{2},\frac{1}{2}\right)$ lies on the normal $x+y-1=0$

Hence, option $\left(A\right)$ is the right answer.