Verified Solution • 2025–26 Session moderate

QJEEM24Ma3162: JEE Main 2024 Mathematics - Class 11 JEEM24Ma3162 Solution: Let the line pass through the point of the intersection (in the …

Let the line L:2x+y=α\mathrm{L}: \sqrt{2} x+y=\alpha pass through the point of the intersection P\mathrm{P} (in the first quadrant) of the circle x2+y2=3x^2+y^2=3 and the parabola x2=2yx^2=2 y . Let the line L\mathrm{L} touch two circles C1\mathrm{C}_1 and C2\mathrm{C}_2 of equal radius 232 \sqrt{3} . If the centres Q1Q_1 and Q2Q_2 of the circles C1C_1 and C2C_2 lie on the yy -axis, then the square of the area of the triangle PQ1Q2\mathrm{PQ}_1 \mathrm{Q}_2 is equal to ___________.

Step-by-Step Verified Solution

x2+y2=3x^2+y^2=3 and x2=2yx^2=2 y

y2+2y−3=0y^2+2 y-3=0

⇒(y+3)(y−1)=0\Rightarrow(y+3)(y-1)=0

y=−3y=-3 (Rejected) or y=1y=1

For y=1,x=2⇒P(2,1)\mathrm{y}=1, \mathrm{x}=\sqrt{2} \Rightarrow P(\sqrt{2}, 1)

pp lies on the line

amp;2x+y=αamp;2(2)+1=αamp;α=3\begin{aligned}& \sqrt{2} x+y=\alpha \\\\& \sqrt{2}(\sqrt{2})+1=\alpha \\\\& \alpha=3\end{aligned}

For circle C1\mathrm{C}_1

Q1\mathrm{Q}_1 lies on y\mathrm{y} axis

Let Q1(0,α)\mathrm{Q}_1(0, \alpha) coordinates

R1=23\mathrm{R}_1=2 \sqrt{3} (Given

Line L\mathrm{L} act as tangent

Apply P =r=r (condition of tangency)

amp;⇒∣α−33∣=23amp;⇒∣α−3∣=6\begin{aligned} & \Rightarrow\left|\frac{\alpha-3}{\sqrt{3}}\right|=2 \sqrt{3} \\\\ & \Rightarrow|\alpha-3|=6\end{aligned}

∴\therefore α−3=6\alpha-3=6

⇒α=9\Rightarrow \alpha=9

 or α−3=−6⇒α=−3\begin{gathered}\text { or } \alpha-3=-6 \\\\ \Rightarrow \alpha=-3\end{gathered}

amp;△PQ1Q2=12∣2amp;1amp;10amp;9amp;10amp;−3amp;1∣amp;=12(2(12))=62amp;(△PQ1Q2)2=72\begin{aligned} & \triangle P Q_1 Q_2=\frac{1}{2}\left|\begin{array}{ccc}\sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1\end{array}\right| \\\\ & =\frac{1}{2}(\sqrt{2}(12))=6 \sqrt{2} \\\\ & \left(\triangle P Q_1 Q_2\right)^2=72\end{aligned}

Related Learning Resources & Exam Preparation