If the straight line `y = 4x+c` is a tangent to the ellipse `x^2//8 + y^2//4 = 1` , then c will be equal to

Options

A. `pm 4`
B. `pm 6`
C. `pmsqrt132`
D. `pm 8`

Answer

To solve the problem, we need to find the value of \( c \) such that the line \( y = 4x + c \) is a tangent to the ellipse given by the equation \( \frac{x^2}{8} + \frac{y^2}{4} = 1 \).

### Step-by-Step Solution:

1. **Identify the parameters of the ellipse:**
The equation of the ellipse is \( \frac{x^2}{8} + \frac{y^2}{4} = 1 \).
Here, we can identify:
- \( a^2 = 8 \) (so \( a = \sqrt{8} = 2\sqrt{2} \))
- \( b^2 = 4 \) (so \( b = \sqrt{4} = 2 \))

2. **Rewrite the line equation:**
The line equation is given as \( y = 4x + c \). We can rewrite it in the standard form \( Ax + By + C = 0 \):
\[ 4x - y + c = 0 \]
Here, \( A = 4 \), \( B = -1 \), and \( C = c \).

3. **Use the condition for tangency:**
For the line to be tangent to the ellipse, the distance from the center of the ellipse (which is at the origin (0,0)) to the line must equal the semi-minor axis \( b \) of the ellipse. The formula for the distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Substituting \( (x_0, y_0) = (0, 0) \):
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|c|}{\sqrt{4^2 + (-1)^2}} = \frac{|c|}{\sqrt{16 + 1}} = \frac{|c|}{\sqrt{17}} \]

4. **Set the distance equal to the semi-minor axis:**
Since the distance must equal \( b = 2 \):
\[ \frac{|c|}{\sqrt{17}} = 2 \]

5. **Solve for \( c \):**
Multiplying both sides by \( \sqrt{17} \):
\[ |c| = 2\sqrt{17} \]
Therefore, \( c \) can take two values:
\[ c = 2\sqrt{17} \quad \text{or} \quad c = -2\sqrt{17} \]

### Final Answer:
Thus, the values of \( c \) are \( c = 2\sqrt{17} \) or \( c = -2\sqrt{17} \).