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QJEEM25Ma66ee: JEE Main 2025 Mathematics - Class 11 JEEM25Ma66ee Solution: If A and B are the points of intersection of the circle and the …

If A and B are the points of intersection of the circle x2+y2−8x=0x^2 + y^2 - 8x = 0 and the hyperbola x29−y24=1\frac{x^2}{9} - \frac{y^2}{4} = 1 and a point P moves on the line 2x−3y+4=02x - 3y + 4 = 0 , then the centroid of ΔPAB\Delta PAB lies on the line :

Multiple Choice Options

  • A x+9y=36x + 9y = 36
  • B 9x−9y=329x - 9y = 32
  • C 4x−9y=124x - 9y = 12
  • D 6x−9y=206x - 9y = 20

Step-by-Step Verified Solution

amp;x2+y2−8x=0,x29−y24=1.... (1)amp;4x2−9y2=36.... (2)amp; Solve (1)&(2)amp;4x2−9(8x−x2)=36amp;13x2−72x−36=0amp;(13x+6)(x=6)=0amp;x=−613,x=6amp;x=−613( rejected )amp;y→ Imaginary amp;n=6,369−y24=1amp;y2=12,y=I12amp;A(6,12),B(6,−12)amp;p(α,2α+43)P lies on \begin{aligned}& x^2+y^2-8 x=0, \frac{x^2}{9}-\frac{y^2}{4}=1 \quad\text{.... (1)}\\& 4 x^2-9 y^2=36 \quad\text{.... (2)}\\& \text { Solve }(1) \&(2) \\& 4 x^2-9\left(8 x-x^2\right)=36 \\& 13 x^2-72 x-36=0 \\& (13 x+6)(x=6)=0 \\& x=\frac{-6}{13}, x=6 \\& x=\frac{-6}{13}(\text { rejected }) \\& y \rightarrow \text { Imaginary } \\& n=6, \frac{36}{9}-\frac{y^2}{4}=1 \\& y^2=12, y=I \sqrt{12} \\& A(6, \sqrt{12}), B(6,-\sqrt{12}) \\& p\left(\alpha, \frac{2 \alpha+4}{3}\right) P \text { lies on }\end{aligned}

amp; centroid (h,k)2x−3y+y=0amp;h=12+α3,α=3 h−12amp;k=2α−3y33⇒2α+4=9yamp;α=9k−42amp;6 h−2y=9k−4amp;6x−9y=20\begin{aligned}& \text { centroid }(\mathrm{h}, \mathrm{k}) \quad 2x-3y+y=0\\& \mathrm{h}=\frac{12+\alpha}{3}, \alpha=3 \mathrm{~h}-12 \\& \mathrm{k}=\frac{\frac{2 \alpha-3 y}{3}}{3} \Rightarrow 2 \alpha+4=9 \mathrm{y} \\& \alpha=\frac{9 \mathrm{k}-4}{2} \\& 6 \mathrm{~h}-2 \mathrm{y}=9 \mathrm{k}-4 \\& 6 \mathrm{x}-9 \mathrm{y}=20\end{aligned}

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