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JEE Advanced 2023 · Conic Sections

Let T1T_1 and T2T_2 be two distinct common tangents to the ellipse E:x26+y23=1E:\frac{x^{2}}{6}+\frac{y^{2}}{3}=1 and the parabola P:y2=12xP:y^{2}=12x. Suppose that the tangent T1T_1 touches PP and EE at the points A1A_1 and A2A_2, respectively and the tangent T2T_2 touches PP and EE at the points A4A_4 and A3A_3, respectively. Then which of the following statements is(are) true?

Answers: A. The area of the quadrilateral A1A2A3A4A_1A_2A_3A_4 is 35 square units · C. The tangents T1T_1 and T2T_2 meet the xx-axis at the point (−3,0)(-3,0)

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Let T₁ and T₂ be two distinct common tangents to the ellipse E:x²/6+y²/3=1 and the parabola P:y²=12x. Suppose that the tangent T₁ touches P and E at the points A₁ and A₂, respectively and the tangent T₂ touches P and E at the points A₄ and A₃, respectively. Then which of the following statements is(are) true? Ellipse E:x26+y23=1\text{Ellipse } E: \dfrac{x^{2}}{6} + \dfrac{y^{2}}{3} = 1 Parabola P:y2=12x\text{Parabola } P: y^{2} = 12x To find: area of quadrilateral A1A2A3A4\text{To find: area of quadrilateral } A_1A_2A_3A_4 and the point where T1,T2 meet the x-axis\text{and the point where } T_1, T_2 \text{ meet the } x\text{-axis} 1. tangent to P with slope m1.\ \text{tangent to } P \text{ with slope } m 2. tangent to E with slope m2.\ \text{tangent to } E \text{ with slope } m 3. common condition⇒m, then the points3.\ \text{common condition} \Rightarrow m,\ \text{then the points} y2=12x=4ax⇒a=3y^{2} = 12x = 4ax \Rightarrow a = 3 T:y=mx+am=mx+3mT: y = mx + \dfrac{a}{m} = mx + \dfrac{3}{m} E:x26+y23=1⇒a2=6, b2=3E: \dfrac{x^{2}}{6} + \dfrac{y^{2}}{3} = 1 \Rightarrow a^{2} = 6,\ b^{2} = 3 y=mx±a2m2+b2=mx±6m2+3y = mx \pm \sqrt{a^{2}m^{2} + b^{2}} = mx \pm \sqrt{6m^{2} + 3} 3m=±6m2+3⇒9m2=6m2+3\dfrac{3}{m} = \pm\sqrt{6m^{2} + 3} \Rightarrow \dfrac{9}{m^{2}} = 6m^{2} + 3 6m4+3m2−9=0⇒2m4+m2−3=06m^{4} + 3m^{2} - 9 = 0 \Rightarrow 2m^{4} + m^{2} - 3 = 0 (2m2+3)(m2−1)=0⇒m2=1⇒m=±1(2m^{2} + 3)(m^{2} - 1) = 0 \Rightarrow m^{2} = 1 \Rightarrow m = \pm 1 T1:y=x+3,T2:y=−x−3T_1: y = x + 3, \qquad T_2: y = -x - 3 T1∩x-axis:x=−3;T2∩x-axis:x=−3T_1 \cap x\text{-axis}: x = -3; \quad T_2 \cap x\text{-axis}: x = -3 ⇒both meet the x-axis at (−3,0)\Rightarrow \text{both meet the } x\text{-axis at } (-3, 0) A1=(at2,2at)=(3,6)(t=1)A_1 = (at^{2}, 2at) = (3, 6) \quad (t = 1) A4=(3,−6)(t=−1)A_4 = (3, -6) \quad (t = -1) A2=(−a2mc,b2c)=(−2,1)A_2 = \left(-\dfrac{a^{2}m}{c}, \dfrac{b^{2}}{c}\right) = (-2, 1) A3=(−2,−1)A_3 = (-2, -1) A2A3=2,A1A4=12,gap=5A_2A_3 = 2, \quad A_1A_4 = 12, \quad \text{gap} = 5 Area=12(A2A3+A1A4)×5=12(2+12)×5=35\text{Area} = \dfrac{1}{2}(A_2A_3 + A_1A_4)\times 5 = \dfrac{1}{2}(2+12)\times 5 = 35 Answer: statements (1) and (3) are true\boxed{\text{Answer: statements (1) and (3) are true}}

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