alleryIf the tangent to the curve $y = e^{2x}$ at point $(a, e^{2a})$ passes through origin, then $a$ equals:
A0
B$1/2$
C1
D$\ln 2$
Answer & Solution
Correct answer: B. $1/2$
Slope at $(a, e^{2a})$: $y' = 2e^{2a}$.
**Tangent:** $y - e^{2a} = 2e^{2a}(x - a)$.
Passes through (0, 0): $-e^{2a} = 2e^{2a}(-a)$
⇒ $1 = 2a$ ⇒ $a = 1/2$.
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