Slope of tangent to `x^(2)=4y` from (-1, -1) can be

Options

A. `(-1+sqrt5)/2`
B. `(-1-sqrt5)/2`
C. `(1-sqrt5)/2`
D. `(1+sqrt5)/2`

Answer

To find the slope of the tangent to the parabola \( x^2 = 4y \) from the point \((-1, -1)\), we can follow these steps:

### Step 1: Understand the parabola
The equation \( x^2 = 4y \) represents a parabola that opens upwards with its vertex at the origin (0,0).

### Step 2: Write the equation of the tangent line
The general equation of a line can be written as:
\[ y = mx + c \]
where \( m \) is the slope and \( c \) is the y-intercept.

### Step 3: Substitute the line equation into the parabola's equation
To find the points of tangency, we substitute \( y = mx + c \) into the parabola's equation:
\[ x^2 = 4(mx + c) \]
This simplifies to:
\[ x^2 - 4mx - 4c = 0 \]

### Step 4: Condition for tangency
For the line to be tangent to the parabola, the quadratic equation must have exactly one solution. This occurs when the discriminant is zero:
\[ b^2 - 4ac = 0 \]
Here, \( a = 1 \), \( b = -4m \), and \( c = -4c \). Thus, we have:
\[ (-4m)^2 - 4(1)(-4c) = 0 \]
This simplifies to:
\[ 16m^2 + 16c = 0 \]
or
\[ c = -m^2 \]

### Step 5: Substitute the point (-1, -1)
Since the line must pass through the point \((-1, -1)\), we substitute these coordinates into the line equation:
\[ -1 = m(-1) + c \]
Substituting \( c = -m^2 \):
\[ -1 = -m + (-m^2) \]
This simplifies to:
\[ m^2 - m - 1 = 0 \]

### Step 6: Solve the quadratic equation for \( m \)
Using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -1 \), and \( c = -1 \):
\[ m = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} \]
\[ m = \frac{1 \pm \sqrt{1 + 4}}{2} \]
\[ m = \frac{1 \pm \sqrt{5}}{2} \]

### Step 7: Final slopes
Thus, the slopes of the tangents to the parabola from the point \((-1, -1)\) can be:
\[ m_1 = \frac{1 + \sqrt{5}}{2}, \quad m_2 = \frac{1 - \sqrt{5}}{2} \]

### Conclusion
The slopes of the tangents to the parabola \( x^2 = 4y \) from the point \((-1, -1)\) are \( \frac{1 + \sqrt{5}}{2} \) and \( \frac{1 - \sqrt{5}}{2} \).

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