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Step-by-Step Solution
Step 1: Identify the Curves
We are given two curves:
1) $y = 10 - x^2$
2) $y = 2 + x^2$
Step 2: Find the Points of Intersection
To find the intersection points, set the two expressions for $y$ equal:
$10 - x^2 = 2 + x^2$
$10 - 2 = x^2 + x^2$
$8 = 2x^2 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = \pm 2
Substitute $x = 2$ into one of the curves (for example, $y = 10 - x^2$):
$y = 10 - (2)^2 = 10 - 4 = 6
So one intersection point is $(2, 6)$.
For $x = -2$:
$y = 10 - (-2)^2 = 10 - 4 = 6
So the other intersection point is $(-2, 6)$.
Step 3: Compute the Slopes of the Tangents to Each Curve
Curve 1: $y = 10 - x^2$
$\frac{dy}{dx} = -2x$
At $x = 2$, slope $m_1 = -2(2) = -4$
At $x = -2$, slope $m_1 = -2(-2) = 4$
Curve 2: $y = 2 + x^2$
$\frac{dy}{dx} = 2x$
At $x = 2$, slope $m_2 = 2(2) = 4$
At $x = -2$, slope $m_2 = 2(-2) = -4$
Step 4: Use the Formula for the Angle between Two Curves
The formula for the tangent of the angle between two curves with slopes $m_1$ and $m_2$ at a point of intersection is:
$\displaystyle \tan \theta \;=\; \left| \frac{m_1 - m_2}{1 + m_1 \, m_2} \right|$
Step 5: Substitute the Slopes into the Formula
Consider the intersection point at $x = 2$:
$m_1 = -4, \quad m_2 = 4$
$\displaystyle \tan \theta \;=\; \left|\frac{-4 - 4}{1 + (-4)(4)}\right|
\;=\; \left|\frac{-8}{1 - 16}\right|
\;=\; \left|\frac{-8}{-15}\right|
\;=\; \frac{8}{15}
The angle remains the same in magnitude for $x = -2$, thus $|\tan \theta| = \frac{8}{15}$.
Step 6: Conclude the Correct Answer
The acute angle between the two curves has
$\displaystyle |\tan \theta| = \frac{8}{15}$.
