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 =-4You need to login to perform this action.
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