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

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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 $\bigg(\large\frac{2 \sqrt 2}{\sqrt 2},\sqrt {\frac{2}{3}}\bigg),\bigg(\large\frac{-2 \sqrt 2}{\sqrt 3},-\sqrt {\frac{2}{3}}\bigg)$
answered Jul 24, 2013 by meena.p
edited Jul 24, 2013 by meena.p
 

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