Question

prove that the evolute if the rectangular hyperbola xy = c^2 is (x+y)^2/3 - (x-y)^2/3 = (4c)^2/3​

Answer 1

Answer:

The evolute of the rectangular hyperbola xy = $c^2$ is $(X + Y)^{2/3 - (X - Y)^{2/3$ = $(4c^2/3)^2$.

Step-by-step explanation:

From the above question,

They have given :

To prove that the evolute if the rectangular hyperbola xy = $c^2 is (x+y)^{2/3 - (x-y)^{2/3$ = $(4c)^{\frac{2}{3} }$

The evolute of a curve is defined as the envelope of the normals to the curve.

To find the evolute of the rectangular hyperbola xy = $c^2$, we can use the fact that the slope of the normal to the hyperbola at a given point (x, y) is equal to -x/y or -y/x, depending on the branch of the hyperbola.

Let (x, y) be a point on the hyperbola xy = $c^2$. Then, the slope of the normal at (x, y) is given by -x/y. Let the normal intersect the evolute at a point (X, Y), and let N be the foot of the perpendicular from (x, y) to the evolute at (X, Y).

Then, we have:

X - x = -y(Y - y)/(X - x) = -y/(-x/y) = - $x^2/y$

Y - y = x(Y - y)/(X - x) = $x/y^2$

Adding these two equations, we get:

X + Y = x(1/$y^2$) - y/(-x/y) = $x^3/y^2 + xy/y$ = $x^2 + c^2$

Similarly, we can obtain the equation X - Y = -x^2 + $c^2$. Hence, we have:

$(X + Y)^2/3 - (X - Y)^{2/3 = [(x^2 + c^2)/3]^2 - [(-x^2 + c^2)/3]^2$

                          = $(4c^2/3)^2$

Thus, we have proved that the evolute of the rectangular hyperbola

         xy = $c^2$ is $(X + Y)^{2/3 - (X - Y)^{2/3$ = $(4c^2/3)^2$.

For more such related questions : https://brainly.in/question/19304141

#SPJ3

Answer 2

Answer:

The evolute of the rectangular hyperbola $xy=c^{2}$  is  $(x+y)^{\frac{2}{3} } -(x-y)^{\frac{2}{3} }=(4c)^{\frac{2}{3} }$.

Step-by-step explanation:

From the above question,

They have given:

To prove that the evolute if the rectangular hyperbola $xy=c^{2}$ is $(x+y)^{\frac{2}{3} } -(x-y)^{\frac{2}{3} }=(4c)^{\frac{2}{3} }$

The evolute of a curve is defined as the envelope of the Normals to the curve.

To find the evolute of the rectangular hyperbola $xy=c^{2}$ , we can use the fact that the slope of the normal to the hyperbola at a given point (x, y) is equal to -x/y or -y/x, depending on the branch of the hyperbola.

Let (x, y) be a point on the hyperbola $xy=c^{2}$ . Then, the slope of the normal at (x, y) is given by -x/y. Let the normal intersect the evolute at a point (X, Y), and let N be the foot of the perpendicular from (x, y) to the evolute at (X, Y).

Then, we have:

X - x = -y(Y - y)/(X - x) = -y/(-x/y) = $-\frac{x^{2} }{y}$

Y - y = x(Y - y)/(X - x) = $\frac{x}{y}$

Adding these two equations, we get:

X + Y = x(1/y^2) - y/(-x/y) =  $\frac{x^{3} }{y^{2} }$$+\frac{xy}{y}$    = $x^{2} +c^{2}$

Similarly, we can obtain the equation X - Y =$-x^{2} +c^{2}$  . Hence, we have:

 $(X+Y)^{2/3} -(X-Y)^{2/3}= (\frac{x^{2} +c^{2}} {y} ) ^{2}-(\frac{-x^{2} +c^{2}} {y} ) ^{2}=(4c)^{2/3}$

Thus, we have proved that the evolute of the rectangular hyperbola

        xy = c^2 is  $(X+Y)^{2/3} -(X-Y)^{2/3}=(4c)^{2/3}$

To know more about hyperbola : https://brainly.com/question/16454195

To know more about slope:  https://brainly.com/question/6531433

#SPJ3