The line 4x+6y+9 =0 touches the parabola `y^(2)=4ax ` at the point

Options

A. `(-3,(9)/(4))`
B. `(3,(-9)/(4))`
C. `((9)/(4),-3)`
D. `((-9)/(4),-3)`

Answer

To solve the problem of finding the point at which the line \(4x + 6y + 9 = 0\) touches the parabola \(y^2 = 4ax\), we can follow these steps:

### Step 1: Write down the equations
The given equations are:
1. Line: \(4x + 6y + 9 = 0\)
2. Parabola: \(y^2 = 4ax\)

### Step 2: Find the slope of the line
To find the slope of the line, we can rearrange the line equation into slope-intercept form \(y = mx + b\):
\[ 6y = -4x - 9 \implies y = -\frac{2}{3}x - \frac{3}{2} \]
Thus, the slope \(m\) of the line is \(-\frac{2}{3}\).

### Step 3: Differentiate the parabola
To find the slope of the tangent to the parabola at any point, we differentiate the parabola equation \(y^2 = 4ax\):
\[ 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]

### Step 4: Set the slopes equal
Since the line is tangent to the parabola, the slopes must be equal at the point of tangency:
\[ -\frac{2}{3} = \frac{2a}{y} \]
Cross-multiplying gives:
\[ -2y = 6a \implies y = -3a \]

### Step 5: Substitute \(y\) back into the parabola equation
Now, substitute \(y = -3a\) into the parabola equation \(y^2 = 4ax\):
\[ (-3a)^2 = 4ax \implies 9a^2 = 4ax \]
Dividing both sides by \(a\) (assuming \(a \neq 0\)):
\[ 9a = 4x \implies x = \frac{9}{4}a \]

### Step 6: Write the point of tangency
Now we have the coordinates of the point of tangency:
\[ (x, y) = \left(\frac{9}{4}a, -3a\right) \]

### Final Answer
The point at which the line touches the parabola is:
\[ \left(\frac{9}{4}a, -3a\right) \]

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