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Parabola question

2025 · 22 Jan · Shift 2 · Q44
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  5. /2025 · 22 Jan · Shift 2 · Q44

Parabola question

2025 · 22 Jan · Shift 2 · Q44

JEE MainMathematicsParabolaMCQ+4 / −1
Let P(4,43)\mathrm{P}(4,4 \sqrt{3})P(4,43​) be a point on the parabola y2=4axy^2=4 \mathrm{a} xy2=4ax 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 :
  1. A
    3433\frac{34 \sqrt{3}}{3}3343​​
  2. B
    34338\frac{343 \sqrt{3}}{8}83433​​
  3. C
    17317 \sqrt{3}173​
  4. D
    26338\frac{263 \sqrt{3}}{8}82633​​
View written solutionFree

Correct answer: B

  1. Find the parabola parameter aaa using point P(4,43)P(4,4\sqrt{3})P(4,43​)

Given parabola: y2=4axy^2=4axy2=4ax Since P(4,43)P(4,4\sqrt{3})P(4,43​) lies on it, (43)2=4a(4)(4\sqrt{3})^2=4a(4)(43​)2=4a(4) 48=16a48=16a48=16a a=3a=3a=3

So the parabola is y2=12xy^2=12xy2=12x with focus (a,0)=(3,0)(a,0)=(3,0)(a,0)=(3,0) and directrix x=−3x=-3x=−3


  1. Parametric form of point PPP

For parabola y2=4axy^2=4axy2=4ax, a point is (at2,2at)(at^2,2at)(at2,2at) Here a=3a=3a=3, so (3t2,6t)=(4,43)(3t^2,6t)=(4,4\sqrt{3})(3t2,6t)=(4,43​) From 6t=436t=4\sqrt{3}6t=43​, t=233t=\frac{2\sqrt{3}}{3}t=323​​ Check: 3t2=3⋅43=43t^2=3\cdot \frac{4}{3}=43t2=3⋅34​=4 correct.

So parameter of PPP is t1=233t_1=\frac{2\sqrt{3}}{3}t1​=323​​


  1. Find the other end QQQ of the focal chord

For parabola y2=4axy^2=4axy2=4ax, if points with parameters t1,t2t_1,t_2t1​,t2​ form a focal chord, then t1t2=−1t_1t_2=-1t1​t2​=−1 Hence t2=−1t1=−32t_2=-\frac{1}{t_1}=-\frac{\sqrt{3}}{2}t2​=−t1​1​=−23​​

Therefore, Q=(3t22,6t2)=(3⋅34,6⋅−32)=(94,−33)Q=(3t_2^2,6t_2)=\left(3\cdot \frac{3}{4},6\cdot -\frac{\sqrt{3}}{2}\right)=\left(\frac{9}{4},-3\sqrt{3}\right)Q=(3t22​,6t2​)=(3⋅43​,6⋅−23​​)=(49​,−33​)


  1. Find points MMM and NNN on the directrix

The directrix is the vertical line x=−3x=-3x=−3 So the foot of perpendicular from a point (x,y)(x,y)(x,y) to the directrix is (−3,y)(-3,y)(−3,y).

Thus, M=(−3,43)M=(-3,4\sqrt{3})M=(−3,43​) N=(−3,−33)N=(-3,-3\sqrt{3})N=(−3,−33​)

So quadrilateral vertices are P(4,43),Q(94,−33),N(−3,−33),M(−3,43)P(4,4\sqrt{3}),\quad Q\left(\frac{9}{4},-3\sqrt{3}\right),\quad N(-3,-3\sqrt{3}),\quad M(-3,4\sqrt{3})P(4,43​),Q(49​,−33​),N(−3,−33​),M(−3,43​)


  1. Compute area of quadrilateral PQMNPQMNPQMN

We use the shoelace formula on vertices in order: P(4,43), Q(94,−33), N(−3,−33), M(−3,43)P(4,4\sqrt{3}),\ Q\left(\frac{9}{4},-3\sqrt{3}\right),\ N(-3,-3\sqrt{3}),\ M(-3,4\sqrt{3})P(4,43​), Q(49​,−33​), N(−3,−33​), M(−3,43​)

Compute S1=4(−33)+94(−33)+(−3)(43)+(−3)(43)S_1=4(-3\sqrt{3})+\frac{9}{4}(-3\sqrt{3})+(-3)(4\sqrt{3})+(-3)(4\sqrt{3})S1​=4(−33​)+49​(−33​)+(−3)(43​)+(−3)(43​) =−123−2734−123−123=-12\sqrt{3}-\frac{27\sqrt{3}}{4}-12\sqrt{3}-12\sqrt{3}=−123​−4273​​−123​−123​ =−17134=-\frac{171\sqrt{3}}{4}=−41713​​

Compute S2=(43)(94)+(−33)(−3)+(−33)(−3)+(43)(4)S_2=(4\sqrt{3})\left(\frac{9}{4}\right)+(-3\sqrt{3})(-3)+(-3\sqrt{3})(-3)+(4\sqrt{3})(4)S2​=(43​)(49​)+(−33​)(−3)+(−33​)(−3)+(43​)(4) =93+93+93+163=9\sqrt{3}+9\sqrt{3}+9\sqrt{3}+16\sqrt{3}=93​+93​+93​+163​ =433=43\sqrt{3}=433​

Area is 12∣S1−S2∣=12∣−17134−433∣\frac12|S_1-S_2|=\frac12\left| -\frac{171\sqrt{3}}{4}-43\sqrt{3} \right|21​∣S1​−S2​∣=21​​−41713​​−433​​ =12∣−1713+17234∣=\frac12\left| -\frac{171\sqrt{3}+172\sqrt{3}}{4} \right|=21​​−41713​+1723​​​ =12⋅34334=\frac12\cdot \frac{343\sqrt{3}}{4}=21​⋅43433​​ =34338=\frac{343\sqrt{3}}{8}=83433​​


  1. Match with options

34338\boxed{\frac{343\sqrt{3}}{8}}83433​​​ So the correct option is B\boxed{\text{B}}B​


  1. Comparison with stored answer

Stored correct answer: B

My derived answer is also B.

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