Question

The equation of the tangent to the curve $x^2-2xy+y^2+2x+y–6=0$ at $(2,2)$ is

• A. $2x+y–6=0$
• B. $x+3y–8=0$
• C. $3x+y–8=0$
• D. $x+y–4=0$

Answer 1

The correct option is A

$2x+y–6=0$

Explanation for the correct option:

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

Given, Equation of curve $x^2-2xy+y^2+2x+y–6=0$

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

$$\begin{gathered} 2x-2\left(y+x\frac{dy}{dx}\right)+2y\frac{dy}{dx}+2+\frac{dy}{dx}=0 \\ \Rightarrow 2x-2y-2x\frac{dy}{dx}+2y\frac{dy}{dx}+2+\frac{dy}{dx}=0 \\ \Rightarrow -2x\frac{dy}{dx}+2y\frac{dy}{dx}+\frac{dy}{dx}=-2x+2y-2 \\ \Rightarrow \frac{dy}{dx}\left(-2x+2y+1\right)=-2x+2y-2 \\ \Rightarrow \frac{dy}{dx}=\frac{-2x+2y-2}{-2x+2y+1} \end{gathered}$$

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

As, the curve passes through the point $(2,2)$

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

$\therefore$Slope of tangent at $(2,2)$ is . ${\left(\frac{dy}{dx}\right)}_{\left(2,2\right)}=m=\frac{-4+4-2}{-4+4+1}$

$m=-2$

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

Therefore, equation of tangent of the curve passing through $(2,2)$ is

$$\begin{gathered} y-2=-2(x-2) \\ \Rightarrow y-2=-2x+4 \\ \Rightarrow 2x+y-6=0 \end{gathered}$$

Hence, the correct answer is option (A).