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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Parabola

Let P(4,4 3 ) be a point on the parabola y^2=4 a x and PQ be a...

Let P(4,43)\mathrm{P}(4,4 \sqrt{3}) be a point on the parabola y2=4axy^2=4 \mathrm{a} x and PQ be a focal chord of the parabola. If M and NN are the foot of perpendiculars drawn from PP and QQ respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to :Let P(4,4 3 ) be a point on the parabola y^2=4 a x and PQ be a focal chord of the parabola. If M and N are the foot of

Options

  1. A17317 \sqrt{3}
  2. B26338\frac{263 \sqrt{3}}{8}
  3. C3433\frac{34 \sqrt{3}}{3}
  4. D34338\frac{343 \sqrt{3}}{8}

Correct answer

D. 34338\frac{343 \sqrt{3}}{8}

Step-by-step solution

Let P(4,4 3 ) be a point on the parabola y^2=4 a x and PQ be a focal chord of the parabola. If M and N are the foot of
(4,43) lies on y2=4ax⇒48=4a⋅44a=12\begin{aligned} & (4,4 \sqrt{3}) \text{ lies on } y^2=4 \mathrm{ax} \\ & \Rightarrow 48=4 \mathrm{a} \cdot 4 \\ & \quad 4 \mathrm{a}=12 \end{aligned} ⇒y2=12x\Rightarrow y^2=12 x is equation of parabola Now, parameter of PP is t1=23⇒t_1=\frac{2}{\sqrt{3}} \Rightarrow Parameters of QQ is t2=−32⇒Q\mathrm{t}_2=-\frac{\sqrt{3}}{2} \Rightarrow \mathrm{Q}\left(\frac{9}{4},-3 \sqrt{3}\right)AreaoftrapeziumPQNMArea of trapezium PQNM=12MN⋅(PM+QN)=12MN⋅(PS+QS)=12MN⋅PQ=1273⋅494=(343)38\begin{aligned} & =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PM}+\mathrm{QN}) \\ & =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PS}+\mathrm{QS}) \\ & =\frac{1}{2} \mathrm{MN} \cdot \mathrm{PQ} \\ & =\frac{1}{2} 7 \sqrt{3} \cdot \frac{49}{4}=(343) \frac{\sqrt{3}}{8} \end{aligned}$
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