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Parabola question

2023 · 13 Apr · Shift 1 · Q24
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  5. /2023 · 13 Apr · Shift 1 · Q24

Parabola question

2023 · 13 Apr · Shift 1 · Q24

JEE MainMathematicsParabolaMCQ+4 / −1
Let PQ\mathrm{PQ}PQ be a focal chord of the parabola y2=36xy^{2}=36 xy2=36x of length 100 , making an acute angle with the positive xxx-axis. Let the ordinate of P\mathrm{P}P be positive and M\mathrm{M}M be the point on the line segment PQ such that PM : MQ = 3 : 1. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line PQ\mathrm{PQ}PQ?
  1. A
    (6,29)(6,29)(6,29)
  2. B
    (−3,43)(-3,43)(−3,43)
  3. C
    (3,33)(3,33)(3,33)
  4. D
    (−6,45)(-6,45)(−6,45)
View written solutionFree

Correct answer: B

  1. Parabola and focal chord parametrization

For the parabola y2=36x,y^2=36x,y2=36x, we compare with the standard form y2=4ax,y^2=4ax,y2=4ax, so 4a=36  ⟹  a=9.4a=36 \implies a=9.4a=36⟹a=9. Hence the focus is S=(9,0).S=(9,0).S=(9,0).

A general point on the parabola is P(at2,2at)=(9t2,18t).P(at^2,2at)=(9t^2,18t).P(at2,2at)=(9t2,18t).

If PQPQPQ is a focal chord of the parabola, then the parameters of its endpoints satisfy t1t2=−1.t_1t_2=-1.t1​t2​=−1. So we may take P=(9t2,18t),Q=(9t2,−18t),P=(9t^2,18t), \qquad Q=\left(\frac{9}{t^2},-\frac{18}{t}\right),P=(9t2,18t),Q=(t29​,−t18​), with t>0t>0t>0 because the ordinate of PPP is positive and the chord makes an acute angle with the positive xxx-axis.

  1. Use the length of the focal chord

For the parabola y2=4axy^2=4axy2=4ax, the length of the focal chord joining parameter points ttt and −1/t-1/t−1/t is a(t+1t)2.a\left(t+\frac{1}{t}\right)^2.a(t+t1​)2. Here this length is given as 100100100, so 9(t+1t)2=100.9\left(t+\frac{1}{t}\right)^2=100.9(t+t1​)2=100. Thus (t+1t)2=1009\left(t+\frac{1}{t}\right)^2=\frac{100}{9}(t+t1​)2=9100​ which gives t+1t=103t+\frac{1}{t}=\frac{10}{3}t+t1​=310​ (since t>0t>0t>0).

Now solve: 3t2−10t+3=03t^2-10t+3=03t2−10t+3=0   ⟹  (3t−1)(t−3)=0.\implies (3t-1)(t-3)=0.⟹(3t−1)(t−3)=0. So t=3ort=13.t=3 \quad \text{or} \quad t=\frac13.t=3ort=31​.

We are told the chord makes an acute angle with the positive xxx-axis. Its slope is

Testing:

  • for t=3t=3t=3, slope is positive,
  • for t=1/3t=1/3t=1/3, slope is also positive but this just swaps endpoints.

Since PPP is the endpoint with positive ordinate, take P=(81,54),Q=(1,−6)P=(81,54),\qquad Q=(1,-6)P=(81,54),Q=(1,−6) (corresponding to t=3t=3t=3).

  1. Find the point MMM dividing PQPQPQ in the ratio PM:MQ=3:1PM:MQ=3:1PM:MQ=3:1

Since PM:MQ=3:1PM:MQ=3:1PM:MQ=3:1, point MMM is closer to QQQ and divides the segment internally in ratio 3:13:13:1. Using section formula, M=(1⋅xP+3⋅xQ4,1⋅yP+3⋅yQ4).M=\left(\frac{1\cdot x_P+3\cdot x_Q}{4},\frac{1\cdot y_P+3\cdot y_Q}{4}\right).M=(41⋅xP​+3⋅xQ​​,41⋅yP​+3⋅yQ​​). So

=\left(\frac{84}{4},\frac{36}{4}\right) =(21,9).$$ --- 4. **Equation of the line through $M$ perpendicular to $PQ$** Slope of $PQ$: $$m_{PQ}=\frac{-6-54}{1-81}=\frac{-60}{-80}=\frac34.$$ Hence the slope of the perpendicular line is $$m_\perp=-\frac43.$$ Equation through $M(21,9)$: $$y-9=-\frac43(x-21).$$ Multiply by 3: $$3y-27=-4x+84$$ $$4x+3y-111=0.$$ So the required line is $$4x+3y=111.$$ --- 5. **Check the given options** We test each point in $$4x+3y=111.$$ ### Option A: $(6,29)$ $$4(6)+3(29)=24+87=111.$$ So A lies on the line. ### Option B: $(-3,43)$ $$4(-3)+3(43)=-12+129=117\neq 111.$$ So B does **not** lie on the line. ### Option C: $(3,33)$ $$4(3)+3(33)=12+99=111.$$ So C lies on the line. ### Option D: $(-6,45)$ $$4(-6)+3(45)=-24+135=111.$$ So D lies on the line. --- 6. **Conclusion** The point that does **not** lie on the required line is $$\boxed{(-3,43)}.$$ So the correct option is **B**.
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