The equation of the tangent to the parabola `y^(2)=8x` inclined at `30^(@)` to the x axis is

Options

A. `3x-sqrt(3)y+4=0`
B. `2x-3y+14=0`
C. `2x-sqrt(2)y+7=0`
D. `x-sqrt(3)y +6=0`

Answer

To find the equation of the tangent to the parabola \( y^2 = 8x \) that is inclined at \( 30^\circ \) to the x-axis, we can follow these steps:

### Step 1: Identify the parameters of the parabola
The given parabola is \( y^2 = 8x \). This can be compared with the standard form \( y^2 = 4ax \). Here, we can see that \( 4a = 8 \), which gives us \( a = 2 \).

**Hint:** Remember that for a parabola in the form \( y^2 = 4ax \), the parameter \( a \) represents the distance from the vertex to the focus.

### Step 2: Determine the slope of the tangent
The slope \( m \) of the tangent line at an angle \( \theta \) to the x-axis is given by \( m = \tan(\theta) \). For \( \theta = 30^\circ \):
\[ m = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

**Hint:** Use the trigonometric identity for tangent to find the slope corresponding to the given angle.

### Step 3: Write the equation of the tangent line
The general equation of the tangent line to the parabola \( y^2 = 4ax \) is given by:
\[ y = mx + c \]
Substituting \( m = \frac{1}{\sqrt{3}} \):
\[ y = \frac{1}{\sqrt{3}}x + c \]

**Hint:** Remember that \( c \) is the y-intercept of the tangent line.

### Step 4: Use the condition of tangency
For the line to be tangent to the parabola, we can use the condition of tangency. The condition states that the discriminant of the quadratic equation formed by substituting the line's equation into the parabola's equation must be zero.

Substituting \( y = \frac{1}{\sqrt{3}}x + c \) into the parabola \( y^2 = 8x \):
\[ \left(\frac{1}{\sqrt{3}}x + c\right)^2 = 8x \]
Expanding this:
\[ \frac{1}{3}x^2 + \frac{2c}{\sqrt{3}}x + c^2 = 8x \]
Rearranging gives:
\[ \frac{1}{3}x^2 + \left(\frac{2c}{\sqrt{3}} - 8\right)x + c^2 = 0 \]

**Hint:** The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by \( D = B^2 - 4AC \).

### Step 5: Set the discriminant to zero
For the line to be tangent, we set the discriminant to zero:
\[ \left(\frac{2c}{\sqrt{3}} - 8\right)^2 - 4 \cdot \frac{1}{3} \cdot c^2 = 0 \]
Simplifying this:
\[ \left(\frac{2c}{\sqrt{3}} - 8\right)^2 = \frac{4}{3}c^2 \]

**Hint:** Solve this equation for \( c \) to find the y-intercept of the tangent line.

### Step 6: Solve for \( c \)
Expanding the left side:
\[ \frac{4c^2}{3} - \frac{32c}{\sqrt{3}} + 64 = \frac{4}{3}c^2 \]
This simplifies to:
\[ -\frac{32c}{\sqrt{3}} + 64 = 0 \]
Solving for \( c \):
\[ \frac{32c}{\sqrt{3}} = 64 \implies c = \frac{64\sqrt{3}}{32} = 2\sqrt{3} \]

**Hint:** Substitute \( c \) back into the tangent line equation.

### Step 7: Write the final equation of the tangent
Substituting \( c = 2\sqrt{3} \) back into the tangent equation:
\[ y = \frac{1}{\sqrt{3}}x + 2\sqrt{3} \]
To write it in standard form:
\[ \sqrt{3}y = x + 6 \implies x - \sqrt{3}y + 6 = 0 \]

Thus, the equation of the tangent to the parabola \( y^2 = 8x \) inclined at \( 30^\circ \) to the x-axis is:
\[ \boxed{x - \sqrt{3}y + 6 = 0} \]