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 \( (0,0) \).

Answer 2

Let's carefully analyze the problem again step-by-step.

Given:
- Parabola: \(y^2 = 8x\)
- Points \(A_1, B_1\) on parabola where lines \(A_1 C_1\) and \(B_1 C_1\) are tangents.
- Point \(C_1 = (-4, 0)\)
- Origin \(O = (0,0)\)

1. Parametric form of points on parabola:
For \(y^2 = 4ax\) with \(a=2\), parametrize as:
\[ x = 2 t^2, \quad y = 4 t \] for parameter \(t\).

2. Tangent at parameter \(t\):
Equation of tangent at point \(t\): \[ y = t x + 2 t^3 \] Or equivalently, \[ t y = x + 2 t^2 \]

3. Point \(C_1 = (-4, 0)\) lies on tangent line:
Substitute \(x=-4, y=0\) in tangent equation: \[ t \cdot 0 = -4 + 2 t^2 \implies 0 = -4 + 2 t^2 \implies 2 t^2 = 4 \implies t^2 = 2 \] So, \[ t = \sqrt{2} \quad \text{or} \quad t = -\sqrt{2} \]

4. Coordinates of \(A_1\) and \(B_1\):
\[ A_1 = (2 t^2, 4 t) = (4, 4 \sqrt{2}) \] \[ B_1 = (4, -4 \sqrt{2}) \]

5. Length \(OA_1\):
\[ OA_1 = \sqrt{(4-0)^2 + (4 \sqrt{2} - 0)^2} = \sqrt{16 + 32} = \sqrt{48} = 4 \sqrt{3} \] So, Statement 1 is TRUE.

6. Check if \(O = (0,0)\) is the orthocenter:
Triangle vertices:
\[ A_1 = (4, 4\sqrt{2}), \quad B_1 = (4, -4\sqrt{2}), \quad C_1 = (-4, 0) \]

Slopes of sides:
- \(A_1 B_1\): vertical line \(x=4\) (slope undefined).
- \(B_1 C_1\): \[ m = \frac{0 + 4\sqrt{2}}{-4 - 4} = \frac{4\sqrt{2}}{-8} = -\frac{\sqrt{2}}{2} \] - \(A_1 C_1\): \[ m = \frac{0 - 4\sqrt{2}}{-4 - 4} = \frac{-4\sqrt{2}}{-8} = \frac{\sqrt{2}}{2} \]

Altitudes:
- Altitude from \(C_1\) is perpendicular to \(A_1 B_1\), so horizontal line through \(C_1\), \(y=0\).
- Altitude from \(B_1\) is perpendicular to \(A_1 C_1\), so slope is negative reciprocal of \(\frac{\sqrt{2}}{2}\) = \(-\sqrt{2}\).
Equation of altitude from \(B_1 (4, -4\sqrt{2})\): \[ y + 4\sqrt{2} = -\sqrt{2}(x - 4) \implies y = -\sqrt{2} x + 4\sqrt{2} - 4\sqrt{2} = -\sqrt{2} x \]

Find intersection of altitudes from \(C_1\) and \(B_1\):
Set \(y=0\) and \(y = -\sqrt{2} x\), so: \[ 0 = -\sqrt{2} x \implies x=0 \] Hence, intersection point is \((0,0) = O\).
This is the orthocenter of \(\triangle A_1 B_1 C_1\).

So, Statement 3 is TRUE.

7. Length \(A_1 B_1\):
\[ A_1 B_1 = \sqrt{(4 - 4)^2 + (4 \sqrt{2} + 4 \sqrt{2})^2} = \sqrt{0 + (8 \sqrt{2})^2} = \sqrt{128} = 8 \sqrt{2} \approx 11.31 \] So Statement 2 (which says 16) is FALSE.

8. Since Statements 1 and 3 are TRUE, and 2 and 4 are FALSE, the correct statements are:

Final Answer:
Statements 1 and 3 are TRUE.