1. Find the point(s) of intersection

The curves are: $y=10-x^2 \quad \text{and} \quad y=2+x^2$

At intersection: $10-x^2 = 2+x^2$ $8 = 2x^2$ $x^2=4$ $x=\pm 2$

Then the corresponding $y$-value is: $y=10-4=6$

So the curves intersect at: $(2,6) \quad \text{and} \quad (-2,6)$

2. Find slopes of the tangents to both curves

For $y=10-x^2$: $\frac{dy}{dx}=-2x$

For $y=2+x^2$: $\frac{dy}{dx}=2x$

At $x=2$:

• Slope of first curve: $m_1=-4$
• Slope of second curve: $m_2=4$

At $x=-2$:

• Slope of first curve: $m_1=4$
• Slope of second curve: $m_2=-4$

In either case, the two tangent slopes are $4$ and $-4$.

3. Use formula for angle between two curves

If $m_1$ and $m_2$ are the slopes of the tangents, then $\tan \theta = \left|\frac{m_2-m_1}{1+m_1m_2}\right|$

Substitute $m_1=4$, $m_2=-4$: $\tan \theta = \left|\frac{-4-4}{1+(4)(-4)}\right|$ $=\left|\frac{-8}{1-16}\right|$ $=\left|\frac{-8}{-15}\right|$ $=\frac{8}{15}$

Thus, $|\tan\theta|=\frac{8}{15}$

4. Check options

• A: $\frac{8}{15}$ ✅
• B: $\frac{4}{9}$ ❌
• C: $\frac{7}{17}$ ❌
• D: $\frac{8}{17}$ ❌

Hence the correct option is A.