Find the locus of the point from which the two tangents drawn to the parabola `y^2=4a x` are such that the slope of one is thrice that of the other.

Explanation

To find the locus of the point from which two tangents are drawn to the parabola \( y^2 = 4ax \), where the slope of one tangent is thrice that of the other, we can follow these steps:

### Step 1: Define the point and slopes
Let the point from which the tangents are drawn be \( P(h, k) \). Let the slopes of the two tangents be \( m_1 \) and \( m_2 \), where it is given that \( m_2 = 3m_1 \).

### Step 2: Write the equation of the tangents
The equation of the tangent to the parabola \( y^2 = 4ax \) at a slope \( m \) is given by:
\[ y = mx + \frac{a}{m} \]

### Step 3: Substitute the point into the tangent equation
Since the tangents pass through the point \( P(h, k) \), we can substitute \( (h, k) \) into the tangent equation:
\[ k = mh + \frac{a}{m} \]
Multiplying through by \( m \) to eliminate the fraction gives:
\[ mk = m^2h + a \]
Rearranging this, we get:
\[ m^2h - mk + a = 0 \]
This is a quadratic equation in \( m \).

### Step 4: Use the condition for the roots
Since \( m_1 \) and \( m_2 \) are the roots of this quadratic equation, we can use Vieta's formulas:
- The sum of the roots \( m_1 + m_2 = \frac{k}{h} \)
- The product of the roots \( m_1 m_2 = \frac{-a}{h} \)

Substituting \( m_2 = 3m_1 \) into the sum gives:
\[ m_1 + 3m_1 = \frac{k}{h} \implies 4m_1 = \frac{k}{h} \implies m_1 = \frac{k}{4h} \]

### Step 5: Substitute \( m_1 \) into the product of the roots
Now substituting \( m_1 = \frac{k}{4h} \) into the product of the roots:
\[ m_1 m_2 = m_1(3m_1) = 3m_1^2 = \frac{-a}{h} \]
Substituting for \( m_1 \):
\[ 3\left(\frac{k}{4h}\right)^2 = \frac{-a}{h} \]
This simplifies to:
\[ \frac{3k^2}{16h^2} = \frac{-a}{h} \]
Multiplying both sides by \( 16h^2 \) gives:
\[ 3k^2 = -16ah \]

### Step 6: Express in terms of \( x \) and \( y \)
Now, replace \( h \) with \( x \) and \( k \) with \( y \):
\[ 3y^2 = -16ax \]
Rearranging gives:
\[ y^2 = \frac{-16a}{3}x \]
This is the equation of a parabola.

### Final Result
Thus, the locus of the point from which the two tangents are drawn is:
\[ y^2 = \frac{-16a}{3}x \]