If the line `y=x+c` touches the hyperbola `(x^(2))/(9)-(y^(2))/(5)=1` at the point `P(h, k),` then `(h, k)` can be equal to

Options

A. `((9)/(2), -(5)/(2))`
B. `((9)/(2),(5)/(2))`
C. `(-(5)/(2),-(9)/(2))`
D. `((5)/(2),-(9)/(2))`

Answer

To solve the problem, we need to find the point \( (h, k) \) where the line \( y = x + c \) touches the hyperbola given by the equation

\[ \frac{x^2}{9} - \frac{y^2}{5} = 1. \]

### Step-by-step Solution:

1. **Identify the parameters of the hyperbola**:
The hyperbola is in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, we can see that \( a^2 = 9 \) and \( b^2 = 5 \). Thus, \( a = 3 \) and \( b = \sqrt{5} \).

2. **Write the equation of the tangent in parametric form**:
The equation of the tangent to the hyperbola in parametric form is given by:
\[ \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1. \]
Substituting the values of \( a \) and \( b \):
\[ \frac{x \sec \theta}{3} - \frac{y \tan \theta}{\sqrt{5}} = 1. \]

3. **Set the tangent equal to the line**:
The line given in the problem is \( y = x + c \). We can rewrite this in the form suitable for comparison:
\[ y - x - c = 0. \]
We need to find the point where this line is tangent to the hyperbola.

4. **Equate the slopes**:
The slope of the line \( y = x + c \) is \( 1 \). The slope of the tangent line from the hyperbola can be found from the parametric form:
\[ \frac{\sqrt{5} \sec \theta}{3 \tan \theta} = 1. \]
This implies:
\[ \sqrt{5} \sec \theta = 3 \tan \theta. \]

5. **Express sec and tan in terms of sin and cos**:
We know that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substituting these into the equation gives:
\[ \sqrt{5} \cdot \frac{1}{\cos \theta} = 3 \cdot \frac{\sin \theta}{\cos \theta}. \]
Cancelling \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ \sqrt{5} = 3 \sin \theta. \]
Thus, we find:
\[ \sin \theta = \frac{\sqrt{5}}{3}. \]

6. **Find the value of \( \cos \theta \)**:
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \left(\frac{\sqrt{5}}{3}\right)^2 + \cos^2 \theta = 1 \implies \frac{5}{9} + \cos^2 \theta = 1 \implies \cos^2 \theta = 1 - \frac{5}{9} = \frac{4}{9}. \]
Therefore,
\[ \cos \theta = \frac{2}{3}. \]

7. **Calculate \( \sec \theta \) and \( \tan \theta \)**:
\[ \sec \theta = \frac{1}{\cos \theta} = \frac{3}{2}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}} = \frac{\sqrt{5}}{2}. \]

8. **Substitute into the tangent equation**:
Now we substitute \( \sec \theta \) and \( \tan \theta \) into the tangent equation:
\[ \frac{x \cdot \frac{3}{2}}{3} - \frac{y \cdot \frac{\sqrt{5}}{2}}{\sqrt{5}} = 1. \]
Simplifying gives:
\[ \frac{x}{2} - \frac{y}{2} = 1 \implies x - y = 2. \]

9. **Find the coordinates of the point of tangency**:
The point of tangency \( (h, k) \) must satisfy both the line \( y = x + c \) and the hyperbola. Substituting \( y = x + c \) into the hyperbola equation:
\[ \frac{x^2}{9} - \frac{(x + c)^2}{5} = 1. \]
This leads to a quadratic equation in \( x \). For the line to be tangent to the hyperbola, the discriminant of this quadratic must be zero.

10. **Solve for \( (h, k) \)**:
After solving, we find that the coordinates of the point of tangency are:
\[ \left( \frac{9}{2}, \frac{5}{2} \right). \]

### Final Answer:
Thus, the point \( (h, k) \) can be equal to \( \left( \frac{9}{2}, \frac{5}{2} \right) \).