Question:

Let A1,B1,C1 A_1, B_1, C_1 be three points in the xy xy -plane. Suppose that the lines A1C1 A_1C_1 and B1C1 B_1C_1 are tangents to the curve y2=8x y^2 = 8x at A1 A_1 and B1 B_1 , respectively. If O=(0,0) O = (0, 0) and C1=(4,0) C_1 = (-4, 0) , then which of the following statements is (are) TRUE?

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For tangents to parabolas, parameterize the points and apply geometric properties of triangles.
Updated On: Jan 20, 2025
  • The length of the line segment OA1 OA_1 is 43 4\sqrt{3} .
  • The length of the line segment A1B1 A_1B_1 is 16 16 .
  • The orthocenter of the triangle A1B1C1 A_1B_1C_1 is (0,0) (0, 0) .
  • The orthocenter of the triangle A1B1C1 A_1B_1C_1 is (1,0) (1, 0) .
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The Correct Option is A

Solution and Explanation

Let A1=(2t12,4t1) A_1 = (2t_1^2, 4t_1) and B1=(2t22,4t2) B_1 = (2t_2^2, 4t_2) . Given C1=(4,0) C_1 = (-4, 0) , we know: t1=2,t2=2. t_1 = -\sqrt{2}, \quad t_2 = \sqrt{2}. Coordinates: A1=(4,42),B1=(4,42). A_1 = (4, -4\sqrt{2}), \quad B_1 = (4, 4\sqrt{2}). Length of OA1 OA_1 : OA1=(40)2+(420)2=43. OA_1 = \sqrt{(4-0)^2 + (-4\sqrt{2}-0)^2} = 4\sqrt{3}. Length of A1B1 A_1B_1 : A1B1=(44)2+(42+42)2=16. A_1B_1 = \sqrt{(4-4)^2 + (4\sqrt{2} + 4\sqrt{2})^2} = 16. The orthocenter of A1B1C1 \triangle A_1B_1C_1 is (0,0) (0, 0) , verified using altitude equations.
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