Question

Let S be the set of all the natural numbers, for which the line
$\frac{x}{a}+\frac{y}{b}=2$
is a tangent to the curve
$(\frac{x}{a})^n+(\frac{y}{b})^n=2$
at the point (a, b), ab ≠ 0. Then :

• A. S=Φ
• B. n(S)=1
• C. S=\left\\{2k:k∈N\right\\}
• D. $S=N$

Answer 1

Answer: (D)

$S=N$

Answer 2

The correct answer is (D) : S=N
$(\frac{x}{a})^n+(\frac{y}{b})^n=2$
$⇒\frac{n}{a}(\frac{x}{a})^{n-1} +\frac{n}{b}(\frac{y}{b})^{n-1}\frac{dy}{dx} =0$
$⇒\frac{dy}{dx}=-\frac{b}{a}(\frac{bx}{ay})^{n-1}$
$\frac{dy}{dx_{(a,b)}}=-\frac{b}{a}$
So line always touches the given curve.