Equation of normal to `x^(2)-4y^(2)=5` at `theta=45^(@)` is

Options

A. `2sqrt2x+2y=5sqrt5`
B. `2sqrt2x+y=5sqrt5`
C. `x+2sqrt2y=5sqrt5`
D. `x-2sqrt2y=5sqrt5`

Answer

To find the equation of the normal to the hyperbola given by \( x^2 - 4y^2 = 5 \) at an angle \( \theta = 45^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Standard Form of the Hyperbola**:
The given equation of the hyperbola is \( x^2 - 4y^2 = 5 \). To write it in standard form, we divide the entire equation by 5:
\[ \frac{x^2}{5} - \frac{4y^2}{5} = 1 \]
This can be rewritten as:
\[ \frac{x^2}{5} - \frac{y^2}{\frac{5}{4}} = 1 \]
Here, \( a^2 = 5 \) and \( b^2 = \frac{5}{4} \).

2. **Determine the Values of \( a \) and \( b \)**:
From the above, we find:
\[ a = \sqrt{5}, \quad b = \frac{\sqrt{5}}{2} \]

3. **Find the Slope of the Normal**:
The slope of the normal line at an angle \( \theta = 45^\circ \) is given by:
\[ m = \tan(45^\circ) = 1 \]

4. **Use the Formula for the Normal Line**:
The equation of the normal to the hyperbola at a point \( (x_0, y_0) \) can be expressed as:
\[ y - y_0 = -\frac{b^2}{a^2}(x - x_0) \]
Since we need the slope of the normal to be 1, we can set:
\[ -\frac{b^2}{a^2} = 1 \]
This gives us:
\[ \frac{b^2}{a^2} = -1 \]
Substituting \( a^2 = 5 \) and \( b^2 = \frac{5}{4} \):
\[ \frac{\frac{5}{4}}{5} = \frac{1}{4} \]

5. **Find the Point of Tangency**:
To find the point of tangency, we can use the parametric equations of the hyperbola:
\[ x = a \sec(t), \quad y = b \tan(t) \]
For \( t = 45^\circ \):
\[ x_0 = \sqrt{5} \sec(45^\circ) = \sqrt{5} \cdot \sqrt{2} = \sqrt{10} \]
\[ y_0 = \frac{\sqrt{5}}{2} \tan(45^\circ) = \frac{\sqrt{5}}{2} \]

6. **Substituting the Point into the Normal Equation**:
Now substituting \( (x_0, y_0) = (\sqrt{10}, \frac{\sqrt{5}}{2}) \) into the normal equation:
\[ y - \frac{\sqrt{5}}{2} = -1(x - \sqrt{10}) \]
Rearranging gives:
\[ y = -x + \sqrt{10} + \frac{\sqrt{5}}{2} \]

7. **Final Equation of the Normal**:
Thus, the equation of the normal to the hyperbola at \( \theta = 45^\circ \) is:
\[ y = -x + \left(\sqrt{10} + \frac{\sqrt{5}}{2}\right) \]