Question

The curve ${\displaystyle {\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n}$ = 2, touches the line ${\displaystyle \frac{x}{a}+\frac{y}{b}}$ = 2 at the point (a, b) for n is equal to ______.

विकल्प

• 1

• 2

• 3

• all non zero values of n

Answer 1

The curve ${\displaystyle {\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n}$ = 2, touches the line ${\displaystyle \frac{x}{a}+\frac{y}{b}}$ = 2 at the point (a, b) for n is equal to all non zero values of n.

Explanation:

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

Differentiating above equation w.r.t.x, we get

${\displaystyle \frac{n{x}^{n-1}}{a^n}+\frac{n{y}^{n-1}}{b^n}\times \frac{dy}{dx}}$ = 0

${\displaystyle \frac{n{y}^{n-1}}{b^n}\times \frac{dy}{dx}=-\frac{n{x}^{n-1}}{a^n}}$

⇒ ${\displaystyle \left(\frac{dy}{dx}\right)=-\frac{n{x}^{n-1}}{a^n}\times \frac{b^n}{n{y}^{n-1}}}$

${\displaystyle \left(\frac{dy}{dx}\right)=-\frac{b^n}{a^n}\times {\left(\frac{x}{y}\right)}^{n-1}}$

${\displaystyle \left(\frac{dy}{dx}\right)=\frac{b^n}{a^n}\times {\left(\frac{a}{b}\right)}^{n-1}}$

= ${\displaystyle -\frac{b^n}{a^n}\times \frac{a^n}{b^n}\times \frac{b}{a}}$

${\displaystyle \frac{dy}{dx}{\mid }_{(a\text{,} b)}=-\frac{b}{a}}$

Compare slope of given tangent slope = ${\displaystyle -\frac{b}{a}}$

Hence, it is valid for all values of n ≠ 0.