The parabola `y^(2)=4x and x^(2)=32y` intersect at a point P other than the origin . If the angle of intersection is `theta` then `tan theta` is equal to

Options

A. `(5)/(3)`
B. `(4)/(5)`
C. `(3)/(5)`
D. 1

Answer

To solve the problem of finding the value of \( \tan \theta \) where \( \theta \) is the angle of intersection of the parabolas \( y^2 = 4x \) and \( x^2 = 32y \), we will follow these steps:

### Step 1: Find the points of intersection
We have two equations:
1. \( y^2 = 4x \)
2. \( x^2 = 32y \)

From the first equation, we can express \( x \) in terms of \( y \):
\[ x = \frac{y^2}{4} \]

Substituting this into the second equation:
\[ \left(\frac{y^2}{4}\right)^2 = 32y \]
\[ \frac{y^4}{16} = 32y \]
Multiplying both sides by 16 to eliminate the fraction:
\[ y^4 = 512y \]
Rearranging gives:
\[ y^4 - 512y = 0 \]
Factoring out \( y \):
\[ y(y^3 - 512) = 0 \]
Thus, \( y = 0 \) or \( y^3 = 512 \). Solving \( y^3 = 512 \) gives:
\[ y = 8 \]

Now substituting \( y = 8 \) back into \( x = \frac{y^2}{4} \):
\[ x = \frac{8^2}{4} = \frac{64}{4} = 16 \]

So the points of intersection are \( (0, 0) \) and \( (16, 8) \).

### Step 2: Find the slopes of the tangents at the point of intersection
**For the parabola \( y^2 = 4x \):**
The slope of the tangent can be found using implicit differentiation:
\[ 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]
At the point \( (16, 8) \):
\[ \frac{dy}{dx} = \frac{2}{8} = \frac{1}{4} \]
Let \( m_1 = \frac{1}{4} \).

**For the parabola \( x^2 = 32y \):**
Using implicit differentiation:
\[ 2x = 32 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{2x}{32} = \frac{x}{16} \]
At the point \( (16, 8) \):
\[ \frac{dy}{dx} = \frac{16}{16} = 1 \]
Let \( m_2 = 1 \).

### Step 3: Find the angle of intersection
The formula for the tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the values of \( m_1 \) and \( m_2 \):
\[ \tan \theta = \left| \frac{\frac{1}{4} - 1}{1 + \frac{1}{4} \cdot 1} \right| \]
Calculating the numerator:
\[ \frac{1}{4} - 1 = \frac{1 - 4}{4} = \frac{-3}{4} \]
Calculating the denominator:
\[ 1 + \frac{1}{4} = \frac{4 + 1}{4} = \frac{5}{4} \]
Thus:
\[ \tan \theta = \left| \frac{-\frac{3}{4}}{\frac{5}{4}} \right| = \left| \frac{-3}{5} \right| = \frac{3}{5} \]

### Final Result
The value of \( \tan \theta \) is:
\[ \tan \theta = \frac{3}{5} \]