Verified Solution • 2025–26 Session moderate

QJEEM25Mae4b6: JEE Main 2025 Mathematics - Class 11 JEEM25Mae4b6 Solution: Let be a point on the parabola and PQ be a focal chord of the…

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 N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to :

Multiple Choice Options

  • A 3433\frac{34 \sqrt{3}}{3}
  • B 34338\frac{343 \sqrt{3}}{8}
  • C 17317 \sqrt{3}
  • D 26338\frac{263 \sqrt{3}}{8}

Step-by-Step Verified Solution

JEE Main 2025 (Online) 22nd January Evening Shift Mathematics - Parabola Question 6 English Explanation

amp;amp;(4,43) lies on y2=4axamp;⇒48=4a.4amp;4a=12amp;⇒y2=12x is equation of parabola amp; Now, parameter of P is t1=23⇒ Parameters of Q\begin{aligned}&\begin{aligned}& (4,4 \sqrt{3}) \text { lies on } \mathrm{y}^2=4 \mathrm{ax} \\& \Rightarrow 48=4 \mathrm{a} .4 \\& \quad 4 \mathrm{a}=12 \\& \Rightarrow \mathrm{y}^2=12 \mathrm{x} \text { is equation of parabola }\end{aligned}\\&\text { Now, parameter of } P \text { is } t_1=\frac{2}{\sqrt{3}} \Rightarrow \text { Parameters of } Q\end{aligned}

amp; is t2=−32⇒Q(94,−33)amp; Area of trapezium PQNM amp;amp;=12MN⋅(PM+QN)amp;=12MN⋅(PS+QS)amp;=12MN⋅PQamp;=1273⋅494=(343)38=3\begin{aligned}&\text { is } \mathrm{t}_2=-\frac{\sqrt{3}}{2} \Rightarrow \mathrm{Q}\left(\frac{9}{4},-3 \sqrt{3}\right)\\&\text { Area of trapezium PQNM }\\&\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}=3\end{aligned}\end{aligned}

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