The sum of the slopes of the tangents to the parabola `y^(2)=8x` drawn from the point (-2,3) is

Options

A. `-2`
B. `-3`
C. `(-3)/(2)`
D. 2

Answer

To find the sum of the slopes of the tangents to the parabola \( y^2 = 8x \) drawn from the point (-2, 3), we can follow these steps:

### Step 1: Identify the parameters of the parabola
The given parabola is \( y^2 = 8x \). We can rewrite this in the standard form \( y^2 = 4ax \). Here, we have:
\[ 4a = 8 \implies a = 2 \]

### Step 2: Write the equation of the tangent line
The equation of the tangent to the parabola \( y^2 = 4ax \) at a point with slope \( m \) is given by:
\[ y = mx + \frac{a}{m} \]
Substituting \( a = 2 \) into the equation, we get:
\[ y = mx + \frac{2}{m} \]

### Step 3: Substitute the point (-2, 3) into the tangent equation
Since the tangent passes through the point (-2, 3), we can substitute \( x = -2 \) and \( y = 3 \) into the tangent equation:
\[ 3 = m(-2) + \frac{2}{m} \]
This simplifies to:
\[ 3 = -2m + \frac{2}{m} \]

### Step 4: Clear the fraction and rearrange the equation
Multiply through by \( m \) to eliminate the fraction:
\[ 3m = -2m^2 + 2 \]
Rearranging gives us:
\[ 2m^2 + 3m - 2 = 0 \]

### Step 5: Solve the quadratic equation
Now we can solve the quadratic equation \( 2m^2 + 3m - 2 = 0 \) using the factorization method or the quadratic formula. Here, we will factor it:
\[ 2m^2 + 4m - m - 2 = 0 \]
Grouping terms, we have:
\[ 2m(m + 2) - 1(m + 2) = 0 \]
Factoring out \( (m + 2) \):
\[ (m + 2)(2m - 1) = 0 \]
Setting each factor to zero gives us:
\[ m + 2 = 0 \quad \text{or} \quad 2m - 1 = 0 \]
Thus, we find:
\[ m = -2 \quad \text{or} \quad m = \frac{1}{2} \]

### Step 6: Calculate the sum of the slopes
Now we sum the slopes:
\[ \text{Sum of slopes} = -2 + \frac{1}{2} = -2 + 0.5 = -\frac{4}{2} + \frac{1}{2} = -\frac{3}{2} \]

### Final Answer
The sum of the slopes of the tangents to the parabola \( y^2 = 8x \) drawn from the point (-2, 3) is:
\[ \boxed{-\frac{3}{2}} \]