Any tangent to a parabola `y^2 = 4ax` and perpendicular to it from the focus meet on the line

Options

A. ` x=0 `
B. `y=0`
C. `x=-a`
D. `y=-a`

Answer

To solve the problem, we need to find the line on which any tangent to the parabola \( y^2 = 4ax \) and a perpendicular line from the focus meet. Let's go through the steps systematically.

### Step 1: Write the equation of the parabola
The given parabola is:
\[ y^2 = 4ax \]

### Step 2: Find the equation of the tangent
The equation of a tangent to the parabola \( y^2 = 4ax \) can be expressed as:
\[ y = mx + \frac{a}{m} \]
where \( m \) is the slope of the tangent.

### Step 3: Rewrite the tangent equation
We can rewrite the tangent equation in standard form:
\[ mx - y = -\frac{a}{m} \]
This is our first equation (let's call it **Equation 1**).

### Step 4: Find the equation of the perpendicular line from the focus
The focus of the parabola \( y^2 = 4ax \) is at the point \( (a, 0) \). The slope of the perpendicular line to the tangent is \( -\frac{1}{m} \). The equation of the line passing through the focus \( (a, 0) \) with slope \( -\frac{1}{m} \) can be written as:
\[ y - 0 = -\frac{1}{m}(x - a) \]
Simplifying this, we get:
\[ y = -\frac{1}{m}x + \frac{a}{m} \]
Rearranging gives us:
\[ \frac{1}{m}x + y = \frac{a}{m} \]
This is our second equation (let's call it **Equation 2**).

### Step 5: Set the equations equal to find the intersection
Now we have:
1. \( mx - y = -\frac{a}{m} \) (Equation 1)
2. \( \frac{1}{m}x + y = \frac{a}{m} \) (Equation 2)

To find the intersection, we can manipulate these equations. From Equation 1, we can express \( y \):
\[ y = mx + \frac{a}{m} \]
Substituting this into Equation 2:
\[ \frac{1}{m}x + (mx + \frac{a}{m}) = \frac{a}{m} \]
Combining terms gives:
\[ \left( \frac{1}{m} + m \right)x + \frac{a}{m} = \frac{a}{m} \]
This simplifies to:
\[ \left( \frac{1}{m} + m \right)x = 0 \]
Thus, we find:
\[ x = 0 \]

### Step 6: Determine the y-coordinate
Since \( x = 0 \), we can substitute back to find \( y \):
Using \( y = mx + \frac{a}{m} \):
\[ y = m(0) + \frac{a}{m} = \frac{a}{m} \]
So the intersection point is \( (0, \frac{a}{m}) \).

### Conclusion
The intersection point lies on the y-axis, confirming that any tangent to the parabola \( y^2 = 4ax \) and the perpendicular line from the focus meet on the line \( x = 0 \), which is the y-axis.