If `t_(1),t_(2),t_(3)` are the feet of normals drawn from `(x_(1),y_(1))` to the parabola `y^(2)=4ax` then the value of `t_(1)t_(2)t_(3)` =

Options

A. a) 0
B. b) `(y_(1))/(a)`
C. c) `(2a-x_(1))/(a)`
D. d) `(x_(1)-2a)/(a)`

Answer

To solve the problem, we need to find the product of the feet of the normals drawn from the point \((x_1, y_1)\) to the parabola given by the equation \(y^2 = 4ax\). Let's go through the solution step by step.

### Step 1: Understand the equation of the normal to the parabola
The equation of the normal to the parabola \(y^2 = 4ax\) at a point where the parameter is \(t\) is given by:
\[ y + tx = 2at + at^3 \]
This can be rearranged to:
\[ at^3 + (2a - x)t - y = 0 \]
This is a cubic equation in \(t\).

**Hint:** Remember that the normal line's equation is derived from the point on the parabola and the slope of the tangent line.

### Step 2: Identify coefficients of the cubic equation
The cubic equation can be compared with the general form \(at^3 + bt^2 + ct + d = 0\). Here:
- \(a = a\) (coefficient of \(t^3\))
- \(b = 0\) (there is no \(t^2\) term)
- \(c = 2a - x\)
- \(d = -y\)

**Hint:** Identify the coefficients carefully to use Vieta's formulas correctly.

### Step 3: Apply Vieta's formulas
According to Vieta's formulas, for a cubic equation \(at^3 + bt^2 + ct + d = 0\), the product of the roots \(t_1, t_2, t_3\) can be expressed as:
\[ t_1 t_2 t_3 = -\frac{d}{a} \]
Substituting our values:
\[ t_1 t_2 t_3 = -\frac{-y}{a} = \frac{y}{a} \]

**Hint:** Vieta's formulas relate the roots of polynomial equations to their coefficients.

### Step 4: Substitute the point \((x_1, y_1)\)
Since we are drawing normals from the point \((x_1, y_1)\), we can substitute \(y_1\) for \(y\):
\[ t_1 t_2 t_3 = \frac{y_1}{a} \]

**Hint:** Ensure that you replace the variable with the correct coordinates from which the normals are drawn.

### Final Answer
Thus, the value of \(t_1 t_2 t_3\) is:
\[ \boxed{\frac{y_1}{a}} \]