Question:
If \(\theta\) denotes the acute angle between the curves, \(y = 10 - x^2\) and \(y = 2 + x^2\) at a point of their intersection, the \(|\tan\theta|\) is equal to :
If \(\theta\) denotes the acute angle between the curves, \(y = 10 - x^2\) and \(y = 2 + x^2\) at a point of their intersection, the \(|\tan\theta|\) is equal to :
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Find the intersection points, the slopes there, and use \(\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|\).
Updated On: Nov 5, 2026
- \(\frac{8}{15}\)
- \(\frac49\)
- \(\frac{7}{17}\)
- \(\frac{8}{17}\)
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The Correct Option is A
Solution and Explanation
Intersection: \(10 - x^2 = 2 + x^2 \Rightarrow x^2 = 4 \Rightarrow x = \pm2,\ y = 6\).
Slopes at \(x = 2\): \(m_1 = -2x = -4\), \(m_2 = 2x = 4\).
\(\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| = \left|\frac{-8}{1 - 16}\right| = \frac{8}{15}\).
(At \(x = -2\) the slopes are \(4\) and \(-4\), giving the same value.)
Answer: (1) \(\frac{8}{15}\)
Slopes at \(x = 2\): \(m_1 = -2x = -4\), \(m_2 = 2x = 4\).
\(\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| = \left|\frac{-8}{1 - 16}\right| = \frac{8}{15}\).
(At \(x = -2\) the slopes are \(4\) and \(-4\), giving the same value.)
Answer: (1) \(\frac{8}{15}\)
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