Question

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:

• A. L touches the curve Γ at the point where the curve crosses the axis of y.
• B. L does not touch the curve at the point where the curve crosses the axis of y.
• C. Γ touches the axis of x at a point.
• D. Γ never touches the axis of x.

Hint:

For intersections of curves and lines, substitute the curve’s equation into the line equation and solve for common points.

Answer 1

The Correct Option is A, D

1. The curve \(\Gamma\) is given by:

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

At \(x = 0\), the curve crosses the \(y\)-axis at \(y = b\).

2. The straight line \(L\) is given by:

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

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

• At \(x = 0\), \(y = b\) for both \(\Gamma\) and \(L\).
• The line \(L\) touches the curve at the point where it crosses the \(y\)-axis.

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

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

Thus, \(L\) touches \(\Gamma\) at the point where \(\Gamma\) crosses the \(y\)-axis.