Question:medium
Let \( \Gamma \) be the curve \( y = b e^{-x/a} \) and \( L \) be the straight line:
\[ \frac{x}{a} + \frac{y}{b} = 1, \quad a, b \in \mathbb{R}. \]
Then:
Let \( \Gamma \) be the curve \( y = b e^{-x/a} \) and \( L \) be the straight line:
\[ \frac{x}{a} + \frac{y}{b} = 1, \quad a, b \in \mathbb{R}. \]
Then:
\[ \frac{x}{a} + \frac{y}{b} = 1, \quad a, b \in \mathbb{R}. \]
Then:
Show Hint
For intersections of curves and lines, substitute the curve’s equation into the line equation and solve for common points.
Updated On: Nov 28, 2025
- L touches the curve Γ at the point where the curve crosses the axis of y.
- L does not touch the curve at the point where the curve crosses the axis of y.
- Γ touches the axis of x at a point.
- Γ never touches the axis of x.
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