WBJEE · Maths · Straight Lines
Let \(\Gamma\) be the curve \(y=b e^{-x / a} \& L\) be the straight line \(\frac{x}{a}+\frac{y}{b}=1\) where \(a, b \in \mathbb{R}\)
- A L touches the curve \(\Gamma\) 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 \(\Gamma\) touches the axis of \(x\) at the point.
- D \(\Gamma\) never touches the axis of \(x\).
Answer & Solution
Correct Answer
(D) \(\Gamma\) never touches the axis of \(x\).
Step-by-step Solution
Detailed explanation
Hint : \(\Gamma \equiv y=b e^{-x / a} \xrightarrow[\text { w.r.t } x]{\text { on differtiating }} \frac{d y}{d x}=\frac{-b}{a} e^{-x / a}\) at \(x=0, y=b, \frac{d y}{d x}=\frac{-b}{a}\) so, equation of tengent at \((0, b)\) is…
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