The slopes of the tangents drawn form (0,2) to the hyperbola `5x^(2)-y^(2)=5` is

Options

A. `3, (-1)/(3)`
B. `-3, 1/3`
C. `pm 2`
D. `pm 3`

Answer

To find the slopes of the tangents drawn from the point (0, 2) to the hyperbola given by the equation \(5x^2 - y^2 = 5\), we can follow these steps:

### Step 1: Rewrite the Hyperbola Equation
First, we rewrite the equation of the hyperbola in standard form by dividing the entire equation by 5:

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

This shows that the hyperbola is centered at the origin (0,0) with \(a^2 = 1\) and \(b^2 = 5\).

### Step 2: Use the Tangent Equation
The equation of the tangent to the hyperbola at a point \((x_1, y_1)\) is given by:

\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]

Here, \(a^2 = 1\) and \(b^2 = 5\). We will substitute \(x_1 = m\) (the slope) and \(y_1 = mx + c\) into the tangent equation.

### Step 3: Substitute the Point (0, 2)
We substitute the point (0, 2) into the tangent equation:

\[ \frac{0 \cdot m}{1} - \frac{2 \cdot (mx + c)}{5} = 1 \]

This simplifies to:

\[ -\frac{2(mx + c)}{5} = 1 \]

### Step 4: Rearranging the Equation
Rearranging gives:

\[ 2(mx + c) = -5 \]

### Step 5: Solve for m
Now, we can express \(c\) in terms of \(m\):

\[ c = -\frac{5}{2} - mx \]

### Step 6: Use the Tangent Condition
For the tangents from the point (0, 2) to the hyperbola, we need to ensure that the discriminant of the resulting quadratic equation in \(m\) is zero. The general form of the tangent line can be expressed as:

\[ y = mx + c \]

Substituting \(y = mx + c\) into the hyperbola equation will yield a quadratic in \(m\).

### Step 7: Set Up the Quadratic Equation
Substituting \(y = mx + c\) into the hyperbola equation gives:

\[ 5x^2 - (mx + c)^2 = 5 \]

Expanding this and rearranging will give us a quadratic in \(m\).

### Step 8: Find the Slopes
We can find the slopes by solving the quadratic equation derived from the substitution. The discriminant must be non-negative for real slopes to exist.

### Step 9: Calculate the Discriminant
The discriminant \(D\) of the quadratic equation must be set to zero to find the slopes:

\[ D = b^2 - 4ac = 0 \]

### Step 10: Solve for m
Solving the quadratic equation will yield the values of \(m\), which are the slopes of the tangents.

### Final Result
After solving, we find that the slopes of the tangents from the point (0, 2) to the hyperbola \(5x^2 - y^2 = 5\) are:

\[ m = \pm 3 \]