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Parabola question

2024 · Shift 2 · Q24
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Parabola question

2024 · Shift 2 · Q24

JEE AdvancedMathematicsParabolaMultiple correct+4 / −2
Let A1,B1,C1A_1, B_1, C_1A1​,B1​,C1​ be three points in the xyx yxy-plane. Suppose that the lines A1C1A_1 C_1A1​C1​ and B1C1B_1 C_1B1​C1​ are tangents to the curve y2=8xy^2=8 xy2=8x at A1A_1A1​ and B1B_1B1​, respectively. If O=(0,0)O=(0,0)O=(0,0) and C1=(−4,0)C_1=(-4,0)C1​=(−4,0), then which of the following statements is (are) TRUE?
  1. A
    The length of the line segment OA1O A_1OA1​ is 434 \sqrt{3}43​
  2. B
    The length of the line segment A1B1A_1 B_1A1​B1​ is 16
  3. C
    The orthocenter of the triangle A1B1C1A_1 B_1 C_1A1​B1​C1​ is (0,0)(0,0)(0,0)
  4. D
    The orthocenter of the triangle A1B1C1A_1 B_1 C_1A1​B1​C1​ is (1,0)(1,0)(1,0)
View written solutionFree

Correct answer: A, C

  1. Parametrize the parabola

For the parabola y2=8x,y^2=8x,y2=8x, we compare with the standard form y2=4ax,y^2=4ax,y2=4ax, so 4a=8  ⟹  a=2.4a=8 \implies a=2.4a=8⟹a=2.

A general point on the parabola is (at2,2at)=(2t2,4t).(at^2,2at)=(2t^2,4t).(at2,2at)=(2t2,4t).

Hence let A1=(2t12,4t1),B1=(2t22,4t2).A_1=(2t_1^2,4t_1),\qquad B_1=(2t_2^2,4t_2).A1​=(2t12​,4t1​),B1​=(2t22​,4t2​).

  1. Equation of tangent at parameter ttt

For y2=4axy^2=4axy2=4ax, the tangent at parameter ttt is ty=x+at2.ty=x+at^2.ty=x+at2. Here a=2a=2a=2, so the tangent is ty=x+2t2.ty=x+2t^2.ty=x+2t2.

Since A1C1A_1C_1A1​C1​ is tangent at A1A_1A1​ and passes through C1=(−4,0)C_1=(-4,0)C1​=(−4,0), substitute (−4,0)(-4,0)(−4,0): t1(0)=−4+2t12t_1(0)=-4+2t_1^2t1​(0)=−4+2t12​ 2t12=42t_1^2=42t12​=4 t12=2t_1^2=2t12​=2 t1=±2.t_1=\pm \sqrt{2}.t1​=±2​.

Similarly, for the tangent at B1B_1B1​ passing through (−4,0)(-4,0)(−4,0), t22=2  ⟹  t2=±2.t_2^2=2 \implies t_2=\pm \sqrt{2}.t22​=2⟹t2​=±2​.

Since there are two distinct tangents from (−4,0)(-4,0)(−4,0) to the parabola, the two contact points correspond to t1=2,t2=−2t_1=\sqrt{2},\qquad t_2=-\sqrt{2}t1​=2​,t2​=−2​ (or vice versa).

Therefore, A1=(2⋅2,42)=(4,42),A_1=(2\cdot 2,4\sqrt{2})=(4,4\sqrt{2}),A1​=(2⋅2,42​)=(4,42​), B1=(2⋅2,−42)=(4,−42).B_1=(2\cdot 2,-4\sqrt{2})=(4,-4\sqrt{2}).B1​=(2⋅2,−42​)=(4,−42​).

  1. Check option A: length OA1OA_1OA1​

Since O=(0,0),A1=(4,42),O=(0,0),\quad A_1=(4,4\sqrt{2}),O=(0,0),A1​=(4,42​), we get

=\sqrt{16+32} =\sqrt{48} =4\sqrt{3}.$$ So **A is true**. --- 4. **Check option B: length $A_1B_1$** The points are $$A_1=(4,4\sqrt{2}),\qquad B_1=(4,-4\sqrt{2}).$$ So $$A_1B_1=|4\sqrt{2}-(-4\sqrt{2})|=8\sqrt{2}.$$ This is not 16. So **B is false**. --- 5. **Check orthocenter of triangle $A_1B_1C_1$** The vertices are $$A_1=(4,4\sqrt{2}),\quad B_1=(4,-4\sqrt{2}),\quad C_1=(-4,0).$$ ### Altitude from $C_1$ Line $A_1B_1$ is vertical: $$x=4.$$ So the altitude from $C_1$ is horizontal: $$y=0.$$ ### Altitude from $A_1$ Slope of $B_1C_1$ is $$m_{B_1C_1}=\frac{0-(-4\sqrt{2})}{-4-4}=\frac{4\sqrt{2}}{-8}=-\frac{\sqrt{2}}{2}.$$ Thus the perpendicular slope is $$m=\sqrt{2}.$$ So altitude from $A_1=(4,4\sqrt{2})$ is $$y-4\sqrt{2}=\sqrt{2}(x-4).$$ At $y=0$, $$-4\sqrt{2}=\sqrt{2}(x-4)$$ $$-4=x-4$$ $$x=0.$$ Thus this altitude meets $y=0$ at $$(0,0).$$ Hence the orthocenter is $$(0,0).$$ So **C is true** and **D is false**. --- 6. **Final conclusion** The true statements are: $$\boxed{A,\ C}$$ This matches the stored correct answer.
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