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JEE Main 201226 MayOnline
Exam
JEE Main
Year
2012
Date
26 May
Mode
Online
Subject
Mathematics
Chapter
Parabola

JEE Main 2012 · 26 May · Online · Mathematics · Parabola

The chord P Q of the parabola y^2=x , where one end P of the chord...

The chord PQP Q of the parabola y2=xy^2=x, where one end PP of the chord is at point (4,−2)(4,-2), is perpendicular to the axis of the parabola. Then the slope of the normal at QQ isThe chord P Q of the parabola y^2=x , where one end P of the chord is at point (4,-2) , is perpendicular to the axis ofThe chord P Q of the parabola y^2=x , where one end P of the chord is at point (4,-2) , is perpendicular to the axis of

Options

  1. A−4-4
  2. B−14-\frac{1}{4}
  3. C4
  4. D14\frac{1}{4}

Correct answer

A. −4-4

Step-by-step solution

Point PP is (4,-2)andPQ⊥xand P Q \perp x-axis So, Q=(4,2)Q=(4,2) The chord P Q of the parabola y^2=x , where one end P of the chord is at point (4,-2) , is perpendicular to the axis of
Equation of tangent at (4,2)(4,2) is yy1=12(x+x1)⇒2y=12(x+2)⇒4y=x+2⇒y=x4+12\begin{aligned}& y y_1=\frac{1}{2}\left(x+x_1\right) \\ & \Rightarrow 2 y=\frac{1}{2}(x+2) \Rightarrow 4 y=x+2 \\ & \Rightarrow y=\frac{x}{4}+\frac{1}{2}\end{aligned}So, slopeoftangent=of tangent =\frac{1}{4}∴\therefore Slope of normal $=-4
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