Q99 / 200 · Manipal (MET) 2014 · Mathematics
The equation of the tangent to the curve $\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2$ at $(a, b)$ is
- Correct answerA$\frac{x}{a}+\frac{y}{b}=2$
- B$\frac{x}{a}+\frac{y}{b}=\frac{1}{2}$
- C$\frac{x}{b}-\frac{y}{a}=2$
- D$a x+b y=2$
Solution
The given curve is $\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2$
$\begin{aligned}
& \Rightarrow n\left(\frac{x}{a}\right)^{n-1} \cdot \frac{1}{a}+n\left(\frac{y}{b}\right)^{n-1} \cdot \frac{1}{b} \frac{d y}{d x}=0 \\
& \Rightarrow \quad \cdot \quad \frac{d y}{d x}=-\frac{b}{a} \frac{\left(\frac{x}{a}\right)^{n-1}}{\left(\frac{y}{b}\right)^{n-1}}
\end{aligned}$
At $(a, b), \frac{d y}{d x}=-\frac{b}{a} \frac{(1)^{n-1}}{(1)^{n-1}}=-\frac{b}{a}$ $\therefore$ The tangent at $(a, b)$ is $\begin{array}{cc} \Rightarrow & y-b=-\frac{b}{a}(x-a) \\ \Rightarrow & a y-a b=-b x+a b \\ \Rightarrow & b x+a y=2 a b \\ \Rightarrow & \frac{x}{a}+\frac{y}{b}=2 \end{array}$
At $(a, b), \frac{d y}{d x}=-\frac{b}{a} \frac{(1)^{n-1}}{(1)^{n-1}}=-\frac{b}{a}$ $\therefore$ The tangent at $(a, b)$ is $\begin{array}{cc} \Rightarrow & y-b=-\frac{b}{a}(x-a) \\ \Rightarrow & a y-a b=-b x+a b \\ \Rightarrow & b x+a y=2 a b \\ \Rightarrow & \frac{x}{a}+\frac{y}{b}=2 \end{array}$
Hence, the correct answer is option: 'A'.