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

$y(x - 2)(x - 3) = x + 6$
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 $\left(\frac{1}{2}, \frac{1}{2}\right)$

Answer 2

The correct option is (A)

(\(\frac{1}{2}\),\(\frac{1}{2}\))

 

Y = \(\frac{x+6}{(x-2)(x-3)}\)

The coordinates of the point of intersection with the y-axis are (0,1)

y=\(\frac{x+6}{x^2-5x+6}\)

⇒y′=(x²−5x+6)−

\(\frac{(2x-5)(x+6))}{(x^{2}-5x+6)^2}\)

⇒y′∣x=0 = 6−\(\frac{(-30)}{36}\)=1

Then, the slope of normal at 

(0,1) is −1

Equation of normal passing through 

(0,1) is y−1=−1(x−0)

I.e., x+y=1

Thus, the normal passes through (\(\frac{1}{2}\),\(\frac{1}{2}\)).