The equation to the common tangent to the parabolas `y^2= 2x` and `x^2 = 16y` is

Options

A. `2x + y +1=0 `
B. `x+2y +2=0`
C. `x+y+3=0`
D. `x-y+4= 0 `

Answer

To find the equation of the common tangent to the parabolas \( y^2 = 2x \) and \( x^2 = 16y \), we can follow these steps:

### Step 1: Identify the equations of the parabolas
The first parabola is given by:
\[ y^2 = 2x \]
This can be rewritten in the standard form as:
\[ x = \frac{y^2}{2} \]

The second parabola is given by:
\[ x^2 = 16y \]
This can be rewritten in the standard form as:
\[ y = \frac{x^2}{16} \]

### Step 2: Write the general equation of the tangent
The general equation of a tangent to the parabola \( y^2 = 2x \) can be expressed as:
\[ y = mx + \frac{a}{m} \]
where \( a \) is a constant.

### Step 3: Find the condition for tangency
For the parabola \( x^2 = 16y \), we can express the tangent line in a similar form:
\[ y = \frac{x^2}{16} + c \]
To find the common tangent, we need to set the two equations equal and find the condition for tangency.

### Step 4: Set up the equations
From the first parabola:
\[ y = mx + \frac{a}{m} \]
Substituting into the second parabola:
\[ mx + \frac{a}{m} = \frac{x^2}{16} \]
Rearranging gives:
\[ x^2 - 16mx - 16\frac{a}{m} = 0 \]

### Step 5: Apply the condition for tangency
For this quadratic equation in \( x \) to have exactly one solution (tangency), the discriminant must be zero:
\[ D = b^2 - 4ac = 0 \]
Here, \( b = -16m \) and \( c = -16\frac{a}{m} \). Thus, we have:
\[ (-16m)^2 - 4(1)(-16\frac{a}{m}) = 0 \]
This simplifies to:
\[ 256m^2 + 64\frac{a}{m} = 0 \]

### Step 6: Solve for \( m \)
Multiplying through by \( m \) (assuming \( m \neq 0 \)):
\[ 256m^3 + 64a = 0 \]
Solving for \( a \):
\[ a = -4m^3 \]

### Step 7: Substitute \( a \) back into the tangent equation
Substituting \( a \) back into the tangent equation:
\[ y = mx - \frac{4m^3}{m} = mx - 4m^2 \]

### Step 8: Rearranging the tangent equation
Rearranging gives:
\[ y = mx - 4m^2 \]
or
\[ mx - y - 4m^2 = 0 \]

### Step 9: Find the specific value of \( m \)
To find the value of \( m \), we can use the condition derived from the tangency condition:
Using the earlier derived condition:
\[ m^2 = -\frac{1}{4} \]
This gives us:
\[ m = -\frac{1}{8} \]

### Step 10: Substitute \( m \) back into the tangent equation
Substituting \( m = -\frac{1}{8} \):
\[ y = -\frac{1}{8}x - 4\left(-\frac{1}{8}\right)^2 \]
Calculating gives:
\[ y = -\frac{1}{8}x - \frac{4}{64} = -\frac{1}{8}x - \frac{1}{16} \]
Rearranging gives:
\[ \frac{1}{8}x + y + \frac{1}{16} = 0 \]
Multiplying through by 16:
\[ 2x + 16y + 1 = 0 \]

### Final Equation
Thus, the equation of the common tangent to the parabolas \( y^2 = 2x \) and \( x^2 = 16y \) is:
\[ 2x + 16y + 1 = 0 \]