A circle drawn on any focal AB of the parabola `y^(2)=4ax` as diameter cute the parabola again at C and D. If the parameters of the points A, B, C, D be `t_(1), t_(2), t_(3)" and "t_(4)` respectively, then the value of `t_(3),t_(4)`, is

Options

A. -1
B. 2
C. 3
D. none of these

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video solution.

### Step 1: Identify the focal points A and B
The parabola given is \( y^2 = 4ax \). The focus of this parabola is located at the point \( (a, 0) \). Since \( AB \) is a focal chord, we can denote the parameters of points \( A \) and \( B \) as \( t_1 \) and \( t_2 \). For a focal chord, the parameters are related by the equation \( t_1 \cdot t_2 = -1 \). Therefore, we can set:
\[ t_1 = -1 \quad \text{and} \quad t_2 = -1 \]

### Step 2: Write the equation of the circle with diameter AB
The coordinates of points \( A \) and \( B \) can be expressed using their parameters:
- Point \( A \) (for \( t_1 \)): \( (at_1^2, 2at_1) = (a(-1)^2, 2a(-1)) = (a, -2a) \)
- Point \( B \) (for \( t_2 \)): \( (at_2^2, 2at_2) = (a(-1)^2, 2a(-1)) = (a, -2a) \)

The center of the circle is the midpoint of \( A \) and \( B \), which is:
\[ \left( \frac{a + a}{2}, \frac{-2a + -2a}{2} \right) = (a, -2a) \]
The radius is half the distance between points \( A \) and \( B \). Since both points are the same, we can use the formula for the circle:
\[ (x - a)^2 + (y + 2a)^2 = r^2 \]

### Step 3: Substitute the parabola's equation into the circle's equation
The equation of the parabola is \( y^2 = 4ax \). We can express \( x \) in terms of \( y \) as:
\[ x = \frac{y^2}{4a} \]
Substituting this into the circle's equation gives:
\[ \left(\frac{y^2}{4a} - a\right)^2 + (y + 2a)^2 = r^2 \]

### Step 4: Expand and simplify the equation
After substituting and simplifying, we will get a fourth-degree polynomial in \( y \). The roots of this polynomial will correspond to the parameters \( t_1, t_2, t_3, t_4 \).

### Step 5: Use the product of the roots
From Vieta's formulas, for a polynomial of the form \( y^4 + py^3 + qy^2 + ry + s = 0 \), the product of the roots \( t_1 t_2 t_3 t_4 = \frac{s}{a} \) (where \( a \) is the leading coefficient). In our case, since \( t_1 = -1 \) and \( t_2 = -1 \):
\[ t_1 t_2 t_3 t_4 = (-1)(-1)(t_3)(t_4) = t_3 t_4 = -3 \]

### Step 6: Solve for \( t_3 \) and \( t_4 \)
Since \( t_1 t_2 = -1 \), we have:
\[ t_3 t_4 = -3 \]
Let \( t_3 = x \) and \( t_4 = y \). Thus, we have:
\[ xy = -3 \]
Given the symmetry and the nature of the parabola, we can assume \( t_3 \) and \( t_4 \) are symmetric about the origin. Therefore, we can set:
\[ t_3 = \sqrt{3}, \quad t_4 = -\sqrt{3} \]
or vice versa.

### Final Answer
Thus, the values of \( t_3 \) and \( t_4 \) are:
\[ t_3 = \sqrt{3}, \quad t_4 = -\sqrt{3} \]