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:

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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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The Correct Option is A, D

Solution and Explanation

1. The curve \(\Gamma\) is defined as:

\[ y = b e^{-x/a}. \]

When \(x = 0\), the curve intersects the \(y\)-axis at \(y = b\).

2. The straight line \(L\) is defined as:

\[ \frac{x}{a} + \frac{y}{b} = 1 \implies y = b \left(1 - \frac{x}{a}\right). \]

3. Analyze the intersection of \(\Gamma\) and \(L\):

  • At \(x = 0\), both \(\Gamma\) and \(L\) have \(y = b\).
  • The line \(L\) is tangent to the curve at the point where the curve intersects the \(y\)-axis.

4. Regarding the \(x\)-axis:

  • As \(x \to \infty\), the curve \(\Gamma\) approaches \(y = 0\), but never touches the \(x\)-axis.

Therefore, \(L\) is tangent to \(\Gamma\) at the \(y\)-axis intersection point.

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