# Find the point on curve $x^{2}-y^{2}=2$ at which the slope of the tangent is $2$

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) Step 1: Let (x_1,y_1) be the point on the curve at which the slope of the tangent is 2 Therefore \large\frac{dy}{dx_{(x_1,y_1)}}$$=2$
Step 2:
Now $x^2-y^2=2$ Differentiating with respect to x
$2x-2y \large\frac{dy}{dx}$$=0 => \large\frac{dy}{dx}=\frac{x}{y} \large\frac{dy}{dx_{(x_1,y_1)}}=\frac{x_1}{y_1}=\frac{x_1}{\sqrt {{x_1}^2-2}}$$=m=-2$
Step 3:
Therefore $\large\frac{x_1}{\sqrt {{x_1}^2-2}}$$=2=>{x_1}^2=4(x_1^2-2) 3{x_1}^2=8=>{x_1}^2=\large\frac{8}{3} =>x_1=\pm \large\frac{2 \sqrt 2}{\sqrt 3} When x_1=\pm \large\frac{2 \sqrt 2}{\sqrt 3}$$ y_1=\sqrt {{x_1}^2-2}=\pm \sqrt {\large\frac{8}{3}-\normalsize 2}=\pm \sqrt {\large\frac{2}{3}}$
Step 4:
Therefore the two points at which the tangent have a slope of $2\bigg(\large\frac{x_1}{y_1}$$>0\bigg)$ are $\bigg(\large\frac{2 \sqrt 2}{\sqrt 2},\sqrt {\frac{2}{3}}\bigg),\bigg(\large\frac{-2 \sqrt 2}{\sqrt 3},-\sqrt {\frac{2}{3}}\bigg)$
edited Jul 24, 2013 by meena.p