Question

The straight line $\frac{x}{a}+\frac{y}{b}=2$ touches the curve ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$ at the point $(a,b)$ for

• A. $n=1,2$
• B. $n=3,4,-5$
• C. $n=1,2,3$
• D. any value of $n$

Answer 1

The correct option is D

any value of $n$

Explanation for the correct option:

Step 1: Compute the required value:

Given the straight line $\frac{x}{a}+\frac{y}{b}=2$

and the curve ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$

Differentiating both sides

$$\begin{gathered} n\frac{x^{n-1}}{a^n}+n\frac{y^{n-1}}{b^n}\frac{dy}{dx}=0 \\ \frac{dy}{dx}\left(\frac{y^{n-1}}{b^n}\right)=-\frac{x^{n-1}}{a^n} \\ \frac{dy}{dx}=-\frac{b^n}{a^n}\left(\frac{x^{n-1}}{y^{n-1}}\right) \end{gathered}$$

Step 2: Compare the slopes of the curve and line

Slope of the curve ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$ at $(a,b)$ is

$$\begin{gathered} \frac{dy}{dx}=-\frac{b^n}{a^n}\left(\frac{a^{n-1}}{b^{n-1}}\right) \\ \frac{dy}{dx}=-\frac{b}{a} \end{gathered}$$

Now, $\frac{x}{a}+\frac{y}{b}=2$

$$\begin{gathered} \Rightarrow \frac{y}{b}=2-\frac{x}{a} \\ \Rightarrow y=2b-\frac{bx}{a} \end{gathered}$$

Thus, its slope$=-\frac{b}{a}$

Since the slope of $\frac{x}{a}+\frac{y}{b}=2$ is also $=-\frac{b}{a}$

So, it touches the curve at $(a,b)$ whatever may be the value of $n$.

Hence, option $\left(D\right)$ is correct.