Point P is (4,−2) and PQ⊥x-axis So, Q=(4,2) 
Equation of tangent at (4,2) is yy1=21(x+x1)⇒2y=21(x+2)⇒4y=x+2⇒y=4x+21 So, slope of tangent =41 ∴ Slope of normal =−4
The chord PQ of the parabola y2=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
Held on 26 May 2012 · Verified 6 Jul 2026.
−4
−41
4
41
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