Let `y^(2)=4ax` be a parabola and `x^(2)-y^(2)=a^(2)` be a hyperbola. Then number of common tangents is

Options

A. 2 for `alt0`
B. 1 for `alt0`
C. 2 for `agt0`
D. 1 for `agt0`

Answer

To find the number of common tangents between the parabola \( y^2 = 4ax \) and the hyperbola \( x^2 - y^2 = a^2 \), we will follow these steps:

### Step 1: Write the equation of the tangent to the parabola
The equation of the tangent to the parabola \( y^2 = 4ax \) can be expressed in the slope-intercept form as:
\[ y = mx + \frac{a}{m} \]
where \( m \) is the slope of the tangent.

**Hint:** Remember that the slope \( m \) can take any real value.

### Step 2: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola \( x^2 - y^2 = a^2 \) is given by:
\[ \frac{x}{a} - \frac{y}{b} = 1 \]
For our hyperbola, we can express it in terms of slope \( m \):
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
where \( b = a \) for our hyperbola, thus:
\[ y = mx \pm \sqrt{a^2 m^2 - a^2} \]
This simplifies to:
\[ y = mx \pm a\sqrt{m^2 - 1} \]

**Hint:** Ensure you understand how to derive the tangent equation for a hyperbola from its standard form.

### Step 3: Set the tangents equal to find common tangents
For the tangents to be common, the equations must be equal:
\[ mx + \frac{a}{m} = mx \pm a\sqrt{m^2 - 1} \]
This leads to two cases:
1. \( \frac{a}{m} = a\sqrt{m^2 - 1} \)
2. \( \frac{a}{m} = -a\sqrt{m^2 - 1} \)

**Hint:** Focus on solving these equations to find the values of \( m \).

### Step 4: Solve the first case
From the first case:
\[ \frac{a}{m} = a\sqrt{m^2 - 1} \]
Assuming \( a \neq 0 \), we can divide both sides by \( a \):
\[ \frac{1}{m} = \sqrt{m^2 - 1} \]
Squaring both sides gives:
\[ \frac{1}{m^2} = m^2 - 1 \]
Multiplying through by \( m^2 \):
\[ 1 = m^4 - m^2 \]
Rearranging gives:
\[ m^4 - m^2 - 1 = 0 \]
Letting \( t = m^2 \), we have:
\[ t^2 - t - 1 = 0 \]
Using the quadratic formula:
\[ t = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]

**Hint:** Use the quadratic formula correctly to find the roots.

### Step 5: Analyze the roots
Since \( t = m^2 \) must be non-negative, we take:
\[ m^2 = \frac{1 + \sqrt{5}}{2} \quad \text{(positive root)} \]
This gives two values for \( m \):
\[ m = \pm \sqrt{\frac{1 + \sqrt{5}}{2}} \]

**Hint:** Remember that each positive value of \( m \) corresponds to two tangents (one for positive and one for negative).

### Step 6: Solve the second case
From the second case:
\[ \frac{a}{m} = -a\sqrt{m^2 - 1} \]
Following similar steps leads to the same quadratic equation, confirming that we have two distinct slopes.

### Conclusion
Thus, we find that there are a total of **2 common tangents** to the parabola and hyperbola, regardless of the value of \( a \).

**Final Answer: 2 common tangents.**