Question

The acute angle between the curves xy = 2, y2 = 4x is​

Answer 1

Answer:

Step-by-step explanation:

tan iy will be the measure of the angle

Answer 2

Answer:

The acute angle between the given curves is​ approximately $71^{o}$.

Step-by-step explanation:

Given the curves $xy = 2$    ...(1)

                    and  $y^{2} = 4x$  ​...(2)

Differentiating equation (1) w.r.t $x$,

                    $xy = 2\implies x\frac{dy}{dx} +y\frac{dx}{dx} =\frac{d(2)}{dx}$

                                 $\implies x\frac{dy}{dx} +y =0$

                                 $\implies xm_{1} +y =0$

where $m_{1}$ is the slope given by, $m_{1} =\frac{-y}{x}$

Similarly, differentiating equation (2) w.r.t $x$,

                    $y^{2} = 4x\implies 2y\frac{dy}{dx} =4\frac{dx}{dx}$

                                 $\implies 2y\frac{dy}{dx} =4$

                                 $\implies 2ym_{2} =4$

where $m_{1}$ is the slope given by, $m_{2} =\frac{4}{2y}=\frac{2}{y}$

Angle between the two curves is given by

                                $\theta=tan^{-1}\Big( \frac{m_{1}- m_{2}}{1+m_{1} m_{2}} \Big)$

Substituting the values of slopes,

                               $\theta=tan^{-1}\Big( \frac{\frac{-y}{x} -\frac{2}{y} }{1+{\frac{-{y}}{x}\frac{2}{y} }} \Big)=tan^{-1}\Big( \frac{\frac{-y^{2}-2x}{xy} }{{\frac{xy-2y}{xy}\ }} \Big)$

                                  $=tan^{-1}\Big( {\frac{-y^{2} -2x}{xy-2y} }{} \Big)$

                                  $=tan^{-1}\Big( {\frac{-y^{2} -2x}{y(x-2)} }{} \Big)$                  ...(3)

To find the values of $x\ \&\ y$, consider the given equations,  

             from $xy = 2\implies x=\frac{2}{y}$                            ...(4)

     and from $y^{2} = 4x\implies x=\frac{y^{2}}{4}$                          ...(5)

Equating equations (4) and (5)      

              $\frac{2}{y}= \frac{y^{2}}{4}\implies y^{3}=8 \implies y=2$

    then value of $x=\frac{2}{y} =\frac{2}{2}=1$

Substituting the values of $x\ \&\ y$ in equation (3),

                          $\theta=tan^{-1}\Big( {\frac{-y^{2} -2x}{y(x-2)} }{} \Big)=tan^{-1}\Big( {\frac{-2^{2} -2\times 1}{2(1-2)} }{} \Big)$

                            $=tan^{-1}( {3 } )$

                            $\approx 71^{o}$

Therefore, the acute angle between the curves is​ approximately $71^{o}$.