The feet of the normals to `y^(2)= 4ax` from the point (6a,0) are

Options

A. (0,0)
B. (4a,-4a)
C. (4a,-4a)
D. (0,0), (4a,4a) ,(4a,-4a)

Answer

To find the feet of the normals to the parabola \( y^2 = 4ax \) from the point \( (6a, 0) \), we can follow these steps:

### Step 1: Understand the equation of the parabola
The given parabola is \( y^2 = 4ax \). This is a standard form of a parabola that opens to the right.

### Step 2: Find the equation of the normal
For a point \( (at^2, 2at) \) on the parabola, the equation of the normal can be derived. The normal at this point is given by:
\[ y - 2at = -\frac{1}{t}(x - at^2) \]
Rearranging this, we get:
\[ y = -\frac{1}{t}x + \left(2at + \frac{at^2}{t}\right) = -\frac{1}{t}x + at + 2at = -\frac{1}{t}x + 3at \]

### Step 3: Substitute the point from which normals are drawn
We need to find the normals from the point \( (6a, 0) \). Substituting \( x = 6a \) and \( y = 0 \) into the normal equation:
\[ 0 = -\frac{1}{t}(6a) + 3at \]
This simplifies to:
\[ \frac{6a}{t} = 3at \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ \frac{6}{t} = 3t \]
Multiplying both sides by \( t \) (assuming \( t \neq 0 \)):
\[ 6 = 3t^2 \]
Thus, we have:
\[ t^2 = 2 \quad \Rightarrow \quad t = \sqrt{2} \text{ or } t = -\sqrt{2} \]

### Step 4: Find the points on the parabola
Now we will find the points on the parabola corresponding to \( t = \sqrt{2} \) and \( t = -\sqrt{2} \):
1. For \( t = \sqrt{2} \):
\[ x = a(\sqrt{2})^2 = 2a, \quad y = 2a(\sqrt{2}) = 2\sqrt{2}a \]
So, the point is \( (2a, 2\sqrt{2}a) \).

2. For \( t = -\sqrt{2} \):
\[ x = a(-\sqrt{2})^2 = 2a, \quad y = 2a(-\sqrt{2}) = -2\sqrt{2}a \]
So, the point is \( (2a, -2\sqrt{2}a) \).

### Step 5: Conclusion
The feet of the normals from the point \( (6a, 0) \) to the parabola \( y^2 = 4ax \) are:
\[ (2a, 2\sqrt{2}a) \quad \text{and} \quad (2a, -2\sqrt{2}a) \]

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