### JEE Main Solved Paper 2017 Question 9

##### 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: [JEE Main Solved Paper-2017]

#### Options:

A) $ ( \frac{1}{2},\frac{1}{3} ) $

B) $ ( -\frac{1}{2},\frac{1}{2} ) $

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

D) $ ( \frac{1}{2},-\frac{1}{3} ) $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

- $ y=\frac{x+6}{(x-2)(x-3)} $ Point of intersection with y-axis (0,1) $ y’=\frac{(x^{2}-5x+6)(1)(x+6)(2x-5)}{{{(x^{2}-5x+6)}^{2}}} $ $ y’=1 $ at point (0,1)

$ \therefore $ Slope of normal is -1 Hence equation of normal is $ x+y=1 $

$ \therefore $ $ ( \frac{1}{2},\frac{1}{2} ) $ satisfy it.