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