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QJEEM23Mabe30: JEE Main 2023 Mathematics - Class 11 JEEM23Mabe30 Solution: Let be a focal chord of the parabola of length 100 , making an…

Let PQ\mathrm{PQ} be a focal chord of the parabola y2=36xy^{2}=36 x of length 100 , making an acute angle with the positive xx -axis. Let the ordinate of P\mathrm{P} be positive and M\mathrm{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} ?

Multiple Choice Options

  • A (6,29)(6,29)
  • B (−3,43)(-3,43)
  • C (3,33)(3,33)
  • D (−6,45)(-6,45)

Step-by-Step Verified Solution

The given parabola is of the form y2=4axy^2 = 4ax . Here, 4a = 36, which means a = 9.

Length of focal chord at (t)=a(t+1t)2=100(t)=a\left(t+\frac{1}{t}\right)^2=100

Where a=9a=9

t+1t=±103∴t=3,13,−3,−13\begin{gathered}t+\frac{1}{t}= \pm \frac{10}{3} \\\\\therefore \quad t=3, \frac{1}{3},-3, \frac{-1}{3}\end{gathered}

Since ordinate of PP is +ve+\mathrm{ve}

∴t=3 or 13\therefore t=3 \text { or } \frac{1}{3}
JEE Main 2023 (Online) 13th April Morning Shift Mathematics - Parabola Question 29 English Explanation

For each value of t, we can find the coordinates of the points P and Q on the parabola using the parametric form of the parabola y2=4axy^2=4ax , where a is 9. The parametric form is given by x=at2x=at^2 and y=2aty=2at .

Let's consider t=3t=3 first.

  1. For t=3t=3 , the coordinates of P are given by P(at2,2at)=P(81,54)P(at^2, 2at) = P(81, 54) .

  2. Since t1t2=−1t_1t_2=-1 for a focal chord, the parameter value for Q is t2=−1/t1=−1/3t_2=-1/t_1=-1/3 . So the coordinates of Q are given by Q(at22,2at2)=Q(1,−6)Q(at_2^2, 2at_2) = Q(1, -6) .

Now, let's find the coordinates of the point M that divides the line segment PQ in the ratio 3:1.

  1. The x-coordinate of M is given by 3Qx+Px4=3∗1+814=21\frac{3Q_x+P_x}{4} = \frac{3*1+81}{4} = 21 .

  2. The y-coordinate of M is given by 3Qy+Py4=3∗(−6)+544=9\frac{3Q_y+P_y}{4} = \frac{3*(-6)+54}{4} = 9 .

So, the coordinates of M are M (21,9)(21, 9) .

We can repeat these steps for t=1/3t=1/3 to find the other set of points P, Q, and M. However, since the ordinate of P is positive and PQ makes an acute angle with the positive x-axis, t=3t=3 is the appropriate choice in this context.

The slope of the line PQ is mPQ=−6−541−81=−60−80=34m_{PQ} = \frac{-6 - 54}{1 - 81} = \frac{-60}{-80} = \frac{3}{4} . The slope of the line perpendicular to PQ is m⊥PQ=−1/mPQ=−43m_{\perp PQ} = -1/m_{PQ} = -\frac{4}{3} . Thus, the equation of the line through M and perpendicular to PQ is (y−9)=−43(x−21)(y - 9) = -\frac{4}{3}(x - 21) , or 4x+3y=1114x + 3y = 111 .

Now, we check which of the given points do NOT lie on this line.

Option A : (6,29)(6, 29) Substitute these values into the equation :

4×6+3×29=111?4\times6 + 3\times29 = 111?

24+87=111?24 + 87 = 111?

111=111111 = 111

So, option A does lie on the line.

Option B : (−3,43)(-3, 43) Substitute these values into the equation :

4×(−3)+3×43=111?4\times(-3) + 3\times43 = 111?

−12+129=111?-12 + 129 = 111?

117=111117 = 111

So, option B does NOT lie on the line.

Option C : (3,33)(3, 33) Substitute these values into the equation :

4×3+3×33=111?4\times3 + 3\times33 = 111?

12+99=111?12 + 99 = 111?

111=111111 = 111

So, option C does lie on the line.

Option D : (−6,45)(-6, 45) Substitute these values into the equation :

4×(−6)+3×45=111?4\times(-6) + 3\times45 = 111?

−24+135=111?-24 + 135 = 111?

111=111111 = 111

So, option D does lie on the line.

Therefore, (−3,43)(-3, 43) which does not lie on the line passing through M and perpendicular to the line PQ.

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