Prove that curve(x/a)n+(y/b)n=2 touches the straight line x/a+y/b=2 at (a,b) for all values of n belongs to N at the - Maths - Application of Derivatives

Question

Prove that curve(x/a)n+(y/b)n=2 touches
the straight line x/a+y/b=2 at (a,b) for all values of n belongs to
N at the point (a,b)

Lalit Mehra · Accepted Answer

Equation of the given curve is $(\frac{x}{a})^{n} + (\frac{y}{b})^{n} = 2$

$(\frac{x}{a})^{n} + (\frac{y}{b})^{n} = 2$

Differentiating both sides w.r.t x, we get

$\frac{n}{a} (\frac{x}{a})^{n - 1} + \frac{n}{b} (\frac{y}{b})^{n - 1} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- \frac{n}{a} (\frac{x}{a})^{n - 1}}{\frac{n}{b} (\frac{y}{b})^{n - 1}} (\frac{d y}{d x})_{(a, b)} = \frac{- b}{a}$

Equation of tangent at (a, b) is

$(y - b) = \frac{- b}{a} (x - a) \Rightarrow a y - a b = - b x + a b \Rightarrow b x + a y = 2 a b \Rightarrow \frac{x}{a} + \frac{y}{b} = 2$

Thus, the given curve touches the line $\frac{x}{a} + \frac{y}{b} = 2$ at (a, b) ∀ n ∈ N.

91

Prove that curve(x/a)n+(y/b)n=2 touches the straight line x/a+y/b=2 at (a,b) for all values of n belongs to N at the point (a,b)..

Prove that curve(x/a)ⁿ+(y/b)ⁿ=2 touches the straight line x/a+y/b=2 at (a,b) for all values of n belongs to N at the point (a,b)..