Let F be the focus of the parabola `y^(2)=4ax` and M be the foot of perpendicular form point `P(at^(2), 2at)` on the tangent at the vertex. If N is a point on the tangent at P, then `(MN)/(FN)"equals"`

Options

A. `t^(2)/(t^(2)+1)`
B. `(t^(2)+1)/t^(2)`
C. 1
D. none of these

Answer

To solve the problem, we need to find the ratio \(\frac{MN}{FN}\) where \(M\) is the foot of the perpendicular from point \(P\) on the tangent at the vertex of the parabola \(y^2 = 4ax\), and \(F\) is the focus of the parabola.

### Step-by-Step Solution:

1. **Identify the Focus and Vertex of the Parabola**:
- The equation of the parabola is \(y^2 = 4ax\).
- The focus \(F\) of the parabola is at the point \((a, 0)\).
- The vertex \(V\) of the parabola is at the origin \((0, 0)\).

**Hint**: Remember that for the parabola \(y^2 = 4ax\), the focus is located at \((a, 0)\) and the vertex is at the origin.

2. **Determine the Coordinates of Point \(P\)**:
- The point \(P\) is given as \(P(at^2, 2at)\).

**Hint**: The coordinates of point \(P\) depend on the parameter \(t\) and follow the form of points on the parabola.

3. **Find the Equation of the Tangent at the Vertex**:
- The equation of the tangent at the vertex (origin) is \(y = 0\) (the x-axis).

**Hint**: The tangent at the vertex of a parabola is horizontal when the parabola opens to the right.

4. **Find the Foot of the Perpendicular \(M\)**:
- The foot of the perpendicular from point \(P(at^2, 2at)\) to the tangent line \(y = 0\) will have the same x-coordinate as \(P\) and a y-coordinate of 0.
- Thus, the coordinates of point \(M\) are \(M(at^2, 0)\).

**Hint**: The foot of the perpendicular to a horizontal line from a point above it will have the same x-coordinate and a y-coordinate of 0.

5. **Calculate the Length \(MN\)**:
- The length \(MN\) is the distance from \(M(at^2, 0)\) to \(N(at^2, 2at)\).
- Using the distance formula:
\[ MN = \sqrt{(at^2 - at^2)^2 + (2at - 0)^2} = \sqrt{(2at)^2} = 2at. \]

**Hint**: The distance between two points with the same x-coordinate is simply the difference in their y-coordinates.

6. **Calculate the Length \(FN\)**:
- The length \(FN\) is the distance from \(F(a, 0)\) to \(N(at^2, 2at)\).
- Using the distance formula:
\[ FN = \sqrt{(at^2 - a)^2 + (2at - 0)^2} = \sqrt{(a(t^2 - 1))^2 + (2at)^2}. \]
- Simplifying:
\[ FN = \sqrt{a^2(t^2 - 1)^2 + 4a^2t^2} = \sqrt{a^2((t^2 - 1)^2 + 4t^2)} = a\sqrt{(t^2 - 1)^2 + 4t^2}. \]

**Hint**: When calculating distances, ensure to square the differences in both x and y coordinates.

7. **Simplify the Expression for \(FN\)**:
- Expanding \((t^2 - 1)^2 + 4t^2\):
\[ (t^2 - 1)^2 + 4t^2 = t^4 - 2t^2 + 1 + 4t^2 = t^4 + 2t^2 + 1 = (t^2 + 1)^2. \]
- Thus, \(FN = a(t^2 + 1)\).

**Hint**: Look for patterns in the expressions to simplify them effectively.

8. **Calculate the Ratio \(\frac{MN}{FN}\)**:
- Now we can find the ratio:
\[ \frac{MN}{FN} = \frac{2at}{a(t^2 + 1)} = \frac{2t}{t^2 + 1}. \]

**Hint**: When finding ratios, ensure to cancel out common factors.

### Final Answer:
\[ \frac{MN}{FN} = \frac{2t}{t^2 + 1}. \]