Question

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

• A. The length of the line segment OA₁ is \(4\sqrt3\)
• B. The length of the line segment A₁B₁ is 16
• C. The orthocenter of the triangle A₁B₁C₁ is (0, 0)
• D. The orthocenter of the triangle A₁B₁C₁ is (1, 0)

Answer 1

The Correct Option is A, C

Triangle Calculation

Let the points be defined as follows: 

• A₁ = \( (2t_1^2, 4t_1) \)
• B₁ = \( (2t_2^2, 4t_2) \)
• C₁ = \( (-4, 0) \)

Given \( t_1 = -\sqrt{2} \) and \( t_2 = \sqrt{2} \), the coordinates are:

• A₁ = \( (4, -4\sqrt{2}) \)
• B₁ = \( (4, 4\sqrt{2}) \)

Step 1: Calculate the length of OA₁

The distance \( OA_1 \) is calculated as:

\[ OA_1 = \sqrt{(4 - 0)^2 + (-4\sqrt{2} - 0)^2} \]

Simplify:

\[ OA_1 = \sqrt{4^2 + (-4\sqrt{2})^2} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3} \]

Step 2: Calculate the length of A₁B₁

The distance \( A_1B_1 \) is given by:

\[ A_1B_1 = \sqrt{(4 - 4)^2 + (4\sqrt{2} - (-4\sqrt{2}))^2} \]

Simplify:

\[ A_1B_1 = \sqrt{0^2 + (8\sqrt{2})^2} = \sqrt{64 \cdot 2} = \sqrt{128} = 16 \]

Step 3: Verify the Orthocenter of △A₁B₁C₁

The orthocenter is the point where the altitudes of the triangle intersect. The altitudes are derived using the slopes of the sides:

• The slope of \( A_1B_1 \) is undefined (vertical line), so its altitude passes through \( C_1 = (-4, 0) \).
• The slopes of \( A_1C_1 \) and \( B_1C_1 \) confirm that the altitudes intersect at the origin \( (0,0) \).

Conclusion:

• \( OA_1 = 4\sqrt{3} \)
• \( A_1B_1 = 16 \)
• The orthocenter is \( (0,0) \).