Question

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

• A. $(b,a)$
• B. $(-b,-a)$
• C. $(a,b)$
• D. none of these

Answer 1

The correct option is C

$(a,b)$

Explanation for correct answer:

Step-1 slope of tangent of the curve:

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

Differentiating both sides

$$\begin{gathered} \Rightarrow \frac{n{x}^{n-1}}{a^n}+\frac{n{y}^{n-1}}{b^n}y\text{'}=0\left[as\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\right] \\ \Rightarrow \frac{n{y}^{n-1}}{b^n}y\text{'}=-\frac{n{x}^{n-1}}{a^n} \\ \Rightarrow y\text{'}=-\frac{b^n{x}^{n-1}}{a^n{y}^{n-1}}------\left(1\right) \end{gathered}$$

Step-2 Slope of tangent :

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

Differentiating both sides

$$\begin{gathered} \Rightarrow \frac{1}{a}+\frac{y\text{'}}{b}=0 \\ \Rightarrow y\text{'}=\frac{-b}{a}----\left(2\right) \end{gathered}$$

Step-3 Point of contact:

From equations (1) and (2)

$$\begin{gathered} -\frac{b}{a}=-\frac{b^n{x}^{n-1}}{a^n{y}^{n-1}} \\ 1=\frac{b^{n-1}{x}^{n-1}}{a^{n-1}{y}^{n-1}} \end{gathered}$$

Taking $n-1$th root both sides

$$\begin{gathered} 1=\frac{bx}{ay} \\ y=\frac{bx}{a} \end{gathered}$$

Put the value of this equation $\frac{x}{a}+\frac{y}{b}=2$

$$\begin{gathered} \frac{x}{a}+\frac{bx}{ab}=2 \\ \frac{x}{a}\left(1+1\right)=2 \\ \frac{x}{a}=1 \\ x=a \end{gathered}$$

$$\begin{gathered} y=\frac{bx}{a} \\ y=\frac{b\times a}{a} \\ y=b \end{gathered}$$

Hence, option C is correct.