if `theta` denotes the acute angle between the curves, `y = 10-x^2" and " y=2+x^2` at a point of their intersection, then `abstantheta` is equal to

Options

A. `4//9`
B. `7//17`
C. `8//17`
D. `8//15`

Answer

To find the value of \( \tan \theta \) where \( \theta \) is the acute angle between the curves \( y = 10 - x^2 \) and \( y = 2 + x^2 \) at their points of intersection, we will follow these steps:

### Step 1: Find the points of intersection of the curves.
We set the two equations equal to each other:
\[ 10 - x^2 = 2 + x^2 \]
Rearranging gives:
\[ 10 - 2 = 2x^2 \implies 8 = 2x^2 \implies x^2 = 4 \implies x = \pm 2 \]

### Step 2: Calculate the corresponding \( y \) values.
Substituting \( x = 2 \) into either equation:
\[ y = 10 - (2)^2 = 10 - 4 = 6 \]
Substituting \( x = -2 \):
\[ y = 10 - (-2)^2 = 10 - 4 = 6 \]
Thus, the points of intersection are \( (2, 6) \) and \( (-2, 6) \).

### Step 3: Find the slopes of the tangents at the points of intersection.
For the curve \( y = 10 - x^2 \):
\[ \frac{dy}{dx} = -2x \]
At \( x = 2 \):
\[ \frac{dy}{dx} = -2(2) = -4 \]
At \( x = -2 \):
\[ \frac{dy}{dx} = -2(-2) = 4 \]

For the curve \( y = 2 + x^2 \):
\[ \frac{dy}{dx} = 2x \]
At \( x = 2 \):
\[ \frac{dy}{dx} = 2(2) = 4 \]
At \( x = -2 \):
\[ \frac{dy}{dx} = 2(-2) = -4 \]

### Step 4: Calculate \( \tan \theta \) using the slopes.
Let \( m_1 = -4 \) (slope of the first curve) and \( m_2 = 4 \) (slope of the second curve). The formula for the tangent of the angle between two curves is:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Substituting the values:
\[ \tan \theta = \frac{-4 - 4}{1 + (-4)(4)} = \frac{-8}{1 - 16} = \frac{-8}{-15} = \frac{8}{15} \]

### Final Result
Thus, \( \tan \theta = \frac{8}{15} \).