The values that m can take so that the straight line `y=4x+m` touches the curve `x^(2)+4y^(2)=4` is

Options

A. `pmsqrt(45)`
B. `pmsqrt(60)`
C. `pmsqrt(65)`
D. `pmsqrt(72)`

Answer

To find the values of \( m \) such that the line \( y = 4x + m \) touches the ellipse given by the equation \( x^2 + 4y^2 = 4 \), we can follow these steps:

### Step 1: Rewrite the ellipse equation
The equation of the ellipse can be rewritten in standard form. We start with:
\[ x^2 + 4y^2 = 4 \]
Dividing through by 4 gives:
\[ \frac{x^2}{4} + \frac{y^2}{1} = 1 \]
This shows that the ellipse has semi-major axis \( a = 2 \) and semi-minor axis \( b = 1 \).

### Step 2: Identify the slope of the tangent line
The line \( y = 4x + m \) has a slope \( m' = 4 \). For the ellipse, the equation of the tangent line at any point can be expressed as:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
where \( m \) is the slope of the tangent line.

### Step 3: Set the slopes equal
From the tangent line equation, we can equate the slopes:
\[ m' = 4 \]
Thus, we can substitute \( m' \) into the tangent line equation:
\[ y = 4x \pm \sqrt{a^2(4^2) + b^2} \]

### Step 4: Substitute the values of \( a \) and \( b \)
Substituting \( a = 2 \) and \( b = 1 \):
\[ \sqrt{a^2(4^2) + b^2} = \sqrt{(2^2)(4^2) + (1^2)} = \sqrt{4 \cdot 16 + 1} = \sqrt{64 + 1} = \sqrt{65} \]

### Step 5: Find the values of \( m \)
Now, substituting back into the tangent line equation:
\[ y = 4x + m \implies m = \pm \sqrt{65} \]
Thus, the values that \( m \) can take are:
\[ m = \sqrt{65} \quad \text{and} \quad m = -\sqrt{65} \]

### Final Answer
The values that \( m \) can take so that the line touches the ellipse are:
\[ m = \sqrt{65} \quad \text{and} \quad m = -\sqrt{65} \]