`y= -2x+12a` is a normal to the parabola `y^(2)=4ax` at the point whose distance from the directrix of the parabola is

Options

A. 4a
B. 5a
C. `4sqrt2a`
D. 8a

Answer

To solve the problem, we need to find the distance from the directrix of the parabola \(y^2 = 4ax\) at the point where the normal line \(y = -2x + 12a\) is tangent to the parabola.

### Step-by-Step Solution:

1. **Identify the Parabola and Directrix**:
The given parabola is \(y^2 = 4ax\). The directrix of this parabola is given by the equation \(x = -a\).

**Hint**: Remember that for a parabola of the form \(y^2 = 4ax\), the directrix is always at \(x = -a\).

2. **Find the Point on the Parabola**:
The normal to the parabola at a point can be expressed in terms of the parameter \(t\). For the parabola \(y^2 = 4ax\), the coordinates of the point on the parabola are given by:
\[ P(t) = (at^2, 2at) \]

**Hint**: The coordinates of a point on the parabola can be derived from the parameter \(t\).

3. **Equation of the Normal**:
The equation of the normal to the parabola at point \(P(t)\) is given by:
\[ y - 2at = -\frac{1}{t}(x - at^2) \]
Rearranging this gives:
\[ y = -\frac{1}{t}x + 2at + \frac{at^2}{t} \]
Simplifying, we get:
\[ y = -\frac{1}{t}x + 2at + at = -\frac{1}{t}x + 3at \]

**Hint**: The slope of the normal line is the negative reciprocal of the slope of the tangent line.

4. **Comparing with the Given Normal**:
We know that the normal line is also given by \(y = -2x + 12a\). By comparing the slopes, we can set:
\[ -\frac{1}{t} = -2 \implies t = \frac{1}{2} \]

**Hint**: Set the slopes equal to find the parameter \(t\).

5. **Find the Coordinates of Point P**:
Substituting \(t = \frac{1}{2}\) into the coordinates of point \(P(t)\):
\[ P\left(\frac{1}{2}\right) = \left(a\left(\frac{1}{2}\right)^2, 2a\left(\frac{1}{2}\right)\right) = \left(\frac{a}{4}, a\right) \]

**Hint**: Substitute the value of \(t\) back into the parameterized equations to find the coordinates.

6. **Calculate the Distance from the Directrix**:
The distance \(d\) from the point \(P\left(\frac{1}{2}\right)\) to the directrix \(x = -a\) is given by:
\[ d = \left| x + a \right| = \left| \frac{a}{4} + a \right| = \left| \frac{a}{4} + \frac{4a}{4} \right| = \left| \frac{5a}{4} \right| = \frac{5a}{4} \]

**Hint**: Use the formula for the distance from a point to a line to find the perpendicular distance.

### Final Answer:
The distance from the directrix of the parabola at the point where the normal intersects is \(\frac{5a}{4}\).