Question

The equation of the tangent to the curve $\left(\frac{x^2}{3}\right)-\left(\frac{y^2}{2}\right)=1$ which is parallel to $y=x$, is

• A. $y=x\pm 1$
• B. $y=x-\left(\frac{1}{2}\right)$
• C. $y=x+\left(\frac{1}{2}\right)$
• D. $y=1-x$

Answer 1

The correct option is A

$y=x\pm 1$

Explantation for correct answer:

Given, $\left(\frac{x^2}{3}\right)-\left(\frac{y^2}{2}\right)=1$

On differentiating with respect to $x$

$\begin{array}{rcl}\frac{2x}{3}-y\left(\frac{dy}{dx}\right)& =& 0\\ & \Rightarrow & \left(\frac{dy}{dx}\right)=\frac{2x}{3y}\end{array}$

Slope of the line $y=xis1.$

Since tangents is parallel to the given line,

So, $\begin{array}{rcl}\left(\frac{2x}{3y}\right)& =& 1\\ & \Rightarrow & x=\frac{3y}{2}\end{array}$

Replacing values we get

$\begin{array}{rcl}\frac{3y^2}{4}-\frac{y^2}{2}& =& 1\\ & \Rightarrow & y^2=4\\ & \Rightarrow & y=\pm 2\end{array}$

The points of contact are $(3,2)and(-3,-2)$

The equations of tangents are $y–2=1(x–3)\Rightarrow x–y–1=0$ and $y+2=1(x+3)\Rightarrow x–y+1=0$

Hence option (A) is the correct answer.