If `x-sqrt(5)y+c=0` is a tangent to the hyperbola `(x^(2))/(25)-(y^(2))/(4)=1` then the value of c is

Options

A. `pm3sqrt(5)`
B. `pmsqrt(5)`
C. `pm2sqrt(5)`
D. none of these

Answer

To find the value of \( c \) such that the line \( x - \sqrt{5}y + c = 0 \) is a tangent to the hyperbola given by \( \frac{x^2}{25} - \frac{y^2}{4} = 1 \), we will follow these steps:

### Step 1: Identify the hyperbola's parameters
The hyperbola is given by the equation:
\[ \frac{x^2}{25} - \frac{y^2}{4} = 1 \]
From this, we can identify:
- \( a^2 = 25 \) (thus \( a = 5 \))
- \( b^2 = 4 \) (thus \( b = 2 \))

### Step 2: Find the slope of the tangent line
The equation of the tangent line can be expressed in the slope-intercept form \( y = mx + c \). We need to find the slope \( m \) of the tangent line. The given line is:
\[ x - \sqrt{5}y + c = 0 \]
Rearranging it gives:
\[ \sqrt{5}y = x + c \quad \Rightarrow \quad y = \frac{1}{\sqrt{5}}x + \frac{c}{\sqrt{5}} \]
Thus, the slope \( m \) is:
\[ m = \frac{1}{\sqrt{5}} \]

### Step 3: Use the tangent condition for hyperbolas
The equation of the tangent to the hyperbola at a point can be given by:
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
Substituting \( m = \frac{1}{\sqrt{5}} \):
\[ y = \frac{1}{\sqrt{5}}x \pm \sqrt{25 \cdot \frac{1}{5} - 4} \]
Calculating the expression under the square root:
\[ 25 \cdot \frac{1}{5} = 5 \quad \Rightarrow \quad 5 - 4 = 1 \]
Thus:
\[ y = \frac{1}{\sqrt{5}}x \pm 1 \]

### Step 4: Compare the two equations
From the tangent line equation:
\[ y = \frac{1}{\sqrt{5}}x + \frac{c}{\sqrt{5}} \]
We can compare this with the tangent line derived from the hyperbola:
\[ y = \frac{1}{\sqrt{5}}x + 1 \quad \text{or} \quad y = \frac{1}{\sqrt{5}}x - 1 \]
This gives us two cases:
1. \(\frac{c}{\sqrt{5}} = 1\) leading to \(c = \sqrt{5}\)
2. \(\frac{c}{\sqrt{5}} = -1\) leading to \(c = -\sqrt{5}\)

### Step 5: Conclusion
Thus, the possible values of \( c \) are:
\[ c = \sqrt{5} \quad \text{or} \quad c = -\sqrt{5} \]

### Final Answer
The value of \( c \) is \( \pm \sqrt{5} \).

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