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).

Answer 1

$$\begin{gathered} \mathrm{Now},{\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2 \\ \frac{n}{a}{\left(\frac{x}{a}\right)}^{n-1}+\frac{n}{b}{\left(\frac{y}{b}\right)}^{n-1}\frac{dy}{dx}=0 \\ \frac{n}{b}{\left(\frac{y}{b}\right)}^{n-1}\frac{dy}{dx}=\frac{-n}{a}{\left(\frac{x}{a}\right)}^{n-1} \\ \frac{dy}{dx}=\frac{-n}{a}{\left(\frac{x}{a}\right)}^{n-1}\times \frac{b}{n}{\left(\frac{b}{y}\right)}^{n-1}=\frac{-b}{a}{\left(\frac{bx}{ay}\right)}^{n-1} \\ \text{Slope of tangent=}{\left(\frac{dy}{dx}\right)}_{\left(a,b\right)}\text{=}\frac{-b}{a}{\left(\frac{b*a}{a*b}\right)}^{n-1}\text{=}\frac{-b}{a}\text{... (2)} \\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{tangent}\mathrm{is} \\ y-b=\frac{-b}{a}\left(x-a\right) \\ \Rightarrow ya-ab=-xb+ab \\ \Rightarrow xb+ya=2ab \\ \Rightarrow \frac{x}{a}+\frac{y}{b}=2 \end{gathered}$$

So, the given line touches the given curve at the given point.