question_answer

The chord PQ of the parabola \[{{y}^{2}}=x,\]where one end P of the chord is at point (4, - 2), is perpendicular to the axis of the parabola. Then the slope of the normal at Q is.   JEE Main Online Paper (Held On 26-May-2012)  

A) \[-4\]    

B)                        \[-\frac{1}{4}\]

C)                         4                                            

D)                        \[\frac{1}{4}\]

Correct Answer: A

Solution :

                Point P is (4, -2) and \[PQ\bot \]x-axis So, Q =(4,2) Equation of tangent at (4, 2) is\[y{{y}_{1}}=\frac{1}{2}(x+{{x}_{1}})\] \[\Rightarrow \]\[2y=\frac{1}{2}(x+2)\Rightarrow 4y=x+2\]\[\Rightarrow \]\[y=\frac{x}{4}+\frac{1}{2}\] So, slope of tangent \[=\frac{1}{4}\] \[\therefore \] Slope of normal =-4