Question:
Two curves \( y = a^x \) and \( y = b^x \) intersect at some angle \( \alpha \). Find \( \tan \alpha \).
Two curves \( y = a^x \) and \( y = b^x \) intersect at some angle \( \alpha \). Find \( \tan \alpha \).
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Use the angle between curves formula involving derivatives to find the tangent of the angle at the point of intersection.
Updated On: Oct 15, 2026
- \( \frac{\log a - \log b}{1 + \log a \log b} \)
- \( \frac{\log a + \log b}{1 - \log a \log b} \)
- \( \frac{\pi}{4} \)
- \( \frac{\pi}{2} \)
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The Correct Option is A
Solution and Explanation
Angle between two curves is given by:
\[
\begin{align}
\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|,\quad \text{where } m_1 = \frac{dy_1}{dx},\ m_2 = \frac{dy_2}{dx}
\]
\[
\begin{align}
\frac{dy}{dx} \text{ for } y = a^x = a^x \log a
\frac{dy}{dx} \text{ for } y = b^x = b^x \log b \] At point of intersection \( a^x = b^x \Rightarrow x = 0 \text{ if } a \ne b \), then: \[ \begin{align} \tan \alpha = \left| \frac{\log a - \log b}{1 + \log a \log b} \right| \]
\frac{dy}{dx} \text{ for } y = b^x = b^x \log b \] At point of intersection \( a^x = b^x \Rightarrow x = 0 \text{ if } a \ne b \), then: \[ \begin{align} \tan \alpha = \left| \frac{\log a - \log b}{1 + \log a \log b} \right| \]
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