BC is latus rectum of a parabola `y^(2)=4ax` and A is its vertex, then minimum length of projection of BC on a tangent drawn in portion BAC is

Options

A. a
B. `2sqrt2a`
C. 2a
D. 3a

Answer

To solve the problem, we need to find the minimum length of the projection of the latus rectum BC of the parabola \(y^2 = 4ax\) on a tangent drawn in the portion BAC.

### Step-by-Step Solution:

1. **Understand the Parabola and its Latus Rectum**:
The given parabola is \(y^2 = 4ax\). The vertex A of the parabola is at the origin (0,0). The latus rectum BC is a line segment that is perpendicular to the axis of the parabola and passes through the focus.

2. **Identify the Coordinates of Points B and C**:
The focus of the parabola \(y^2 = 4ax\) is at the point (a,0). The endpoints of the latus rectum (B and C) are located at:
- \(B(a, 2a)\)
- \(C(a, -2a)\)

3. **Length of the Latus Rectum**:
The length of the latus rectum BC is the distance between points B and C:
\[ BC = |2a - (-2a)| = |2a + 2a| = 4a \]

4. **Finding the Tangent at Point A**:
The equation of the tangent to the parabola at point A (0,0) can be expressed as:
\[ y = mx \]
where \(m\) is the slope of the tangent.

5. **Angle of Projection**:
The angle \(\theta\) between the latus rectum BC (which is vertical) and the tangent line can be calculated using the slope \(m\):
\[ \tan \theta = m \]

6. **Projection of BC on the Tangent**:
The projection of the length of BC on the tangent line is given by:
\[ \text{Projection} = BC \cdot \cos \theta \]
Since \(BC = 4a\), we have:
\[ \text{Projection} = 4a \cdot \cos \theta \]

7. **Finding \(\cos \theta\)**:
Using the relationship between the tangent and cosine:
\[ \cos \theta = \frac{1}{\sqrt{1 + m^2}} \]
Thus, the projection becomes:
\[ \text{Projection} = 4a \cdot \frac{1}{\sqrt{1 + m^2}} \]

8. **Minimizing the Projection**:
To find the minimum length of the projection, we need to minimize the expression \(4a \cdot \frac{1}{\sqrt{1 + m^2}}\). The minimum occurs when \(m = 1\) (the tangent is at a 45-degree angle):
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \]
Therefore, the minimum projection is:
\[ \text{Minimum Projection} = 4a \cdot \frac{1}{\sqrt{2}} = 2\sqrt{2}a \]

### Final Answer:
The minimum length of the projection of BC on a tangent drawn in the portion BAC is \(2\sqrt{2}a\).

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