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Question
Prove that \[\left( \frac{x}{a} \right)^n + \left( \frac{y}{b} \right)^n = 2\] touches the straight line \[\frac{x}{a} + \frac{y}{b} = 2\] for all n ∈ N, at the point (a, b) ?
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Solution
\[\text { Now }, \left( \frac{x}{a} \right)^n + \left( \frac{y}{b} \right)^n = 2\]
\[\frac{n}{a} \left( \frac{x}{a} \right)^{n - 1} + \frac{n}{b} \left( \frac{y}{b} \right)^{n - 1} \frac{dy}{dx} = 0\]
\[\frac{n}{b} \left( \frac{y}{b} \right)^{n - 1} \frac{dy}{dx} = \frac{- n}{a} \left( \frac{x}{a} \right)^{n - 1} \]
\[\frac{dy}{dx} = \frac{- n}{a} \left( \frac{x}{a} \right)^{n - 1} \times \frac{b}{n} \left( \frac{b}{y} \right)^{n - 1} = \frac{- b}{a} \left( \frac{bx}{ay} \right)^{n - 1} \]
\[\text { Slope of tangent }= \left( \frac{dy}{dx} \right)_\left( a, b \right) =\frac{- b}{a} \left( \frac{b * a}{a * b} \right)^{n - 1} =\frac{- b}{a}... (2)\]
\[\text { The equation of tangent is }\]
\[y - b = \frac{- b}{a}\left( x - a \right)\]
\[ \Rightarrow ya - ab = - xb + ab\]
\[ \Rightarrow xb + ya = 2ab\]
\[ \Rightarrow \frac{x}{a} + \frac{y}{b} = 2\]
So, the given line touches the given curve at the given point.
