Assertion(A) : If the line x = 3y+k touches the parabola `3y^(2) = 4x` then k = 5
Reason (R) : Equation to the tangent `y^(2)= 8x` inclimed at an angle `30^(@)` to the axis is x-`sqrt(3)y+6=0`

Options

A. Both A and R are true and R `rarr`A
B. Both A and R are true and `R cancelimplies a`
C. A is true but R is false
D. A is false but R is true

Answer

To solve the problem, we will analyze the assertion and the reason step by step.

### Step 1: Understanding the Assertion
The assertion states that if the line \( x = 3y + k \) touches the parabola \( 3y^2 = 4x \), then \( k = 5 \).

### Step 2: Rewrite the Line Equation
We can rewrite the line equation in standard form:
\[ x - 3y - k = 0 \]
Here, \( l = 1 \), \( m = -3 \), and \( n = -k \).

### Step 3: Determine the Parabola's Standard Form
The equation of the parabola is given as \( 3y^2 = 4x \). We can rewrite this in the standard form:
\[ y^2 = \frac{4}{3}x \]
From this, we can identify \( a \) (the distance from the vertex to the focus) as:
\[ 4a = \frac{4}{3} \implies a = \frac{1}{3} \]

### Step 4: Condition for Tangency
The condition for the line \( lx + my + n = 0 \) to be a tangent to the parabola \( y^2 = 4ax \) is given by:
\[ ln = am^2 \]
Substituting the values we have:
\[ (1)(-k) = \left(\frac{1}{3}\right)(-3)^2 \]
This simplifies to:
\[ -k = \frac{1}{3} \cdot 9 \implies -k = 3 \implies k = -3 \]

### Step 5: Conclusion for Assertion
Since we found \( k = -3 \), which contradicts the assertion that \( k = 5 \), we conclude that the assertion is **false**.

### Step 6: Understanding the Reason
The reason states that the equation of the tangent \( y^2 = 8x \) is inclined at an angle of \( 30^\circ \) to the axis, given by the line \( x - \sqrt{3}y + 6 = 0 \).

### Step 7: Finding the Slope from the Reason
From the line equation \( x - \sqrt{3}y + 6 = 0 \), we can express it in slope-intercept form:
\[ \sqrt{3}y = x + 6 \implies y = \frac{1}{\sqrt{3}}x + 2\sqrt{3} \]
Thus, the slope \( m \) of the line is:
\[ m = \frac{1}{\sqrt{3}} \]

### Step 8: Angle of Inclination
The angle of inclination \( \theta \) is given by:
\[ \tan \theta = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ \]
This confirms that the tangent line is inclined at \( 30^\circ \) to the x-axis.

### Step 9: Conclusion for Reason
Since the slope from the line matches the expected slope for an angle of \( 30^\circ \), the reason is **true**.

### Final Conclusion
- Assertion (A) is **false**.
- Reason (R) is **true**.