Given the ellipse `(x ^(2))/(4) +y ^(2) =1,` the point on the line `x =2,` such that the tangents drawn from the point to the ellipse are at `45^(@),` is

Options

A. `(2,2 + sqrt5)`
B. `(2 sqrt5)`
C. `(2, sqrt3)`
D. `(2, -sqrt3)`

Answer

To solve the problem, we need to find the point on the line \( x = 2 \) such that the tangents drawn from this point to the ellipse \( \frac{x^2}{4} + y^2 = 1 \) are at an angle of \( 45^\circ \).

### Step-by-Step Solution:

1. **Identify the Ellipse Parameters**:
The given ellipse is \( \frac{x^2}{4} + y^2 = 1 \). This can be rewritten in standard form as:
\[ \frac{x^2}{2^2} + \frac{y^2}{1^2} = 1 \]
Here, \( a = 2 \) and \( b = 1 \).

2. **Point on the Line**:
We need to find a point on the line \( x = 2 \). Thus, the point can be represented as \( (2, y) \), where \( y \) is variable.

3. **Tangent Condition**:
The tangents from the point \( (2, y) \) to the ellipse should make an angle of \( 45^\circ \) with the x-axis. The slope of the tangent line at this angle is:
\[ m = \tan(45^\circ) = 1 \]

4. **Equation of the Tangent**:
The equation of the tangent to the ellipse at point \( (x_0, y_0) \) is given by:
\[ \frac{xx_0}{4} + yy_0 = 1 \]
Substituting \( x_0 = 2 \) into the tangent equation gives:
\[ \frac{2x}{4} + yy_0 = 1 \implies \frac{x}{2} + yy_0 = 1 \]

5. **Finding the y-coordinate**:
Since the slope \( m = 1 \), we can write the equation of the tangent line through the point \( (2, y) \) as:
\[ y - y_0 = 1(x - 2) \implies y = x - 2 + y_0 \]
Setting \( y_0 = y \) gives:
\[ y = x - 2 + y \implies x - 2 = 0 \implies x = 2 \]

6. **Using the Tangent Condition**:
The condition for the tangent line to be at \( 45^\circ \) is:
\[ c = \pm \sqrt{m^2(a^2 + b^2)} \]
where \( m = 1 \), \( a = 2 \), and \( b = 1 \):
\[ c = \pm \sqrt{1^2(2^2 + 1^2)} = \pm \sqrt{1(4 + 1)} = \pm \sqrt{5} \]

7. **Finding y-values**:
The y-coordinate can be calculated as:
\[ y = 2 + c \quad \text{or} \quad y = 2 - c \]
Therefore, we have:
\[ y = 2 + \sqrt{5} \quad \text{and} \quad y = 2 - \sqrt{5} \]

8. **Identifying Valid Points**:
The points we get are:
\[ (2, 2 + \sqrt{5}) \quad \text{and} \quad (2, 2 - \sqrt{5}) \]
Since \( \sqrt{5} \approx 2.236 \), the point \( (2, 2 - \sqrt{5}) \) will have a negative y-coordinate, which means it lies inside the ellipse. Therefore, it is not valid.

9. **Final Answer**:
The valid point from which the tangents can be drawn at \( 45^\circ \) is:
\[ \boxed{(2, 2 + \sqrt{5})} \]