Question

The equation of the tangent to the curve $x=\frac{t-1}{t+1}$, $y=\frac{t+1}{t-1}$ at $t=2$ is

• A. $x+9y–6=0$
• B. $9x–y–6=0$
• C. $9x+y+6=0$
• D. $x+9y+6=0$
• E. $9x+y–6=0$

Answer 1

The correct option is E

$9x+y–6=0$

Explanation for the correct option:

Step 1: Find the first order derivative of the equation of the given curve wrt $x$

Given, equation of curve $x=\frac{t-1}{t+1}...\left(i\right)$ and $y=\frac{t+1}{t-1}...\left(ii\right)$

On multiplying eq $\left(\mathrm{i}\right)$ by eq $\left(\mathrm{ii}\right)$ we get,

$$\begin{gathered} xy=1 \\ \Rightarrow y=\frac{1}{x} \end{gathered}$$

On differentiating w.r.t. $x$, we get

$\frac{dy}{dx}=\frac{-1}{x^2}$

Step 2:Find the slope of the tangent at the required point

Now at $t=2$,

$x=\frac{2-1}{2+1}andy=\frac{2+1}{2-1}$

$x=\frac{1}{3}$ and $y=3$ [from eq $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ respectively]

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

$\therefore$Slope of tangent at $\left(\frac{1}{3},3\right)$ is ${\left(\frac{dy}{dx}\right)}_{\left(\frac{1}{3},3\right)}=-9$

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

Therefore, equation of tangent of the curve passing through $\left(\frac{1}{3},3\right)$ is

$$\begin{gathered} y-3=-9\left(x-\frac{1}{3}\right) \\ \Rightarrow y-3=-9x+\frac{9}{3} \\ \Rightarrow 9x+y-6=0 \end{gathered}$$

Hence, the correct answer is option (E).