Question

The curve $\frac{x^n}{a^n}+\frac{y^n}{b^n}=2$ touches the line $\frac{x}{a}+\frac{y}{b}=2$ at the point

• A. $(b,a)$
• B. $(a,b)$
• C. $(1,1)$
• D. $(\frac{1}{a},\frac{1}{b})$

Answer 1

Answer: (B)

$(a,b)$

Slope of curve$\frac{x^n}{a^n}+\frac{y^n}{b^n}=2$ is
$\frac{dy}{dx}=-\frac{b^n}{a^n}\frac{x^{n-1}}{y^{n-1}}$
Slope of tangent $\frac{x}{a}+\frac{y}{b}=2$ is
$\frac{dy}{dx}=-\frac{b}{a}$
Equating both slopes, we get
$-\frac{b^n}{a^n}\frac{x^{n-1}}{y^{n-1}}=-\frac{b}{a}$
Only $x=ay=b$ satisfy the above equation.
Hence, option 'B' is correct.