Question

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

• A. $\left[\frac{x}{a}\right]+\left[\frac{y}{b}\right]=2$
• B. $\left[\frac{x}{a}\right]+\left[\frac{y}{b}\right]=\frac{1}{2}$
• C. $\left[\frac{x}{a}\right]-\left[\frac{y}{b}\right]=2$
• D. $ax+by=2$

Answer 1

The correct option is A

$\left[\frac{x}{a}\right]+\left[\frac{y}{b}\right]=2$

Explanation for the correct answer:

Step1: Find first order derivative of given function

Given, equation of curve ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$

On differentiating the above equation w.r.t. $x$, we get

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

Step 2: Find slope of tangent at given point

The first order derivative at a given point gives the slope of the tangent to the curve at the given point

$\therefore$Slope of tangent at $\left(a,b\right)$ is ${\left(\frac{dy}{dx}\right)}_{\left(a,b\right)}=m=\frac{-b}{a}{\left(\frac{a}{a}\right)}^{n-1}{\left(\frac{b}{b}\right)}^{n-1}$

$m=\frac{-b}{a}$

Step 3: Find the equation of the tangent using slope-point form of line

The equation of a line in slope-point form can be written as

$\mathbf{y}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}\mathbf{=}\mathbf{m}\left(x-x_1\right)$

where $m$ is the slope of the line and $\left(x_1,y_1\right)$ are the co-ordinates of the point

Therefore, equation of tangent to the curve passing through $\left(a,b\right)$ is

$$\begin{gathered} y-b=-\frac{b}{a}\left(x-a\right) \\ \Rightarrow y-b=\frac{-bx}{a}+b \\ \Rightarrow y-b-b+\frac{bx}{a}=0 \\ \Rightarrow \frac{ay-2ab+bx}{a}=0 \\ \Rightarrow ay-2ab+bx=0 \\ \Rightarrow bx+ay=2ab \\ \Rightarrow \frac{x}{a}+\frac{y}{b}=2 \end{gathered}$$

Hence, the correct answer is option (A).