The condition that the line y=mx+c may be a tangent to `(y^(2))/(a^(2))-x^(2)/b^(2)=1` is

Options

A. `c^(2)=a^(2)m^(2)-b^(2)`
B. `c^(2)=a^(2)-b^(2)m^2`
C. `c^2=b^(2)-a^(2)m^(2)`
D. `c^(2)=b^(2)m^(2)-a^(2)`

Answer

To find the condition that the line \( y = mx + c \) may be a tangent to the hyperbola given by the equation

\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1, \]

we will follow these steps:

### Step 1: Rewrite the hyperbola equation
The hyperbola can be rewritten in standard form as:

\[ y^2 = a^2 + \frac{a^2}{b^2} x^2. \]

### Step 2: Substitute the line equation into the hyperbola
Substituting \( y = mx + c \) into the hyperbola equation gives:

\[ \frac{(mx + c)^2}{a^2} - \frac{x^2}{b^2} = 1. \]

### Step 3: Expand the equation
Expanding the left-hand side:

\[ \frac{m^2x^2 + 2mcx + c^2}{a^2} - \frac{x^2}{b^2} = 1. \]

### Step 4: Combine terms
To combine the terms, we first find a common denominator:

\[ \frac{m^2x^2 + 2mcx + c^2}{a^2} - \frac{a^2x^2}{a^2b^2} = 1. \]

This leads to:

\[ \frac{(m^2 - \frac{a^2}{b^2})x^2 + \frac{2mc}{a^2}x + \frac{c^2}{a^2} - 1}{a^2} = 0. \]

### Step 5: Set the quadratic equation
This gives us a quadratic equation in terms of \( x \):

\[ (m^2 - \frac{a^2}{b^2})x^2 + \frac{2mc}{a^2}x + \left(\frac{c^2}{a^2} - 1\right) = 0. \]

### Step 6: Use the condition for tangency
For the line to be a tangent to the hyperbola, the discriminant of this quadratic equation must be zero:

\[ \left(\frac{2mc}{a^2}\right)^2 - 4(m^2 - \frac{a^2}{b^2})\left(\frac{c^2}{a^2} - 1\right) = 0. \]

### Step 7: Simplify the discriminant
Expanding the discriminant gives:

\[ \frac{4m^2c^2}{a^4} - 4(m^2 - \frac{a^2}{b^2})(\frac{c^2}{a^2} - 1) = 0. \]

### Step 8: Rearranging terms
Rearranging and simplifying leads to the condition:

\[ c^2 = a^2m^2 - b^2. \]

### Final Result
Thus, the condition that the line \( y = mx + c \) may be a tangent to the hyperbola is:

\[ c^2 = a^2m^2 - b^2. \]