Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/232

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Find the rectangular equation of the envelope of the straight line $$y = mx + \frac{p}{m}$$, where the slope $$m$$ is the variable parameter.


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 * style="width:10%"|Solution.
 * align="center" |$$y = mx + \frac{p}{m}$$
 * }


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 * style="width:10%"|First Step.
 * align="center" |$$0 = x - \frac{p}{m^2}$$.
 * }


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 * style="width:10%"|Solving,
 * align="center" |$$m = \pm \sqrt{\frac{p}{x}}$$.
 * }




 * Substitute in the given equation,

$y = \pm \sqrt{\frac{p}{x}} \cdot x \pm \sqrt{\frac{x}{p}} \cdot p = \pm 2 \sqrt{px}$.|undefined


 * and squaring, $$y^2 = 4px$$, a parabola, is the equation of the envelope. The family of straight lines formed by varying the slope $$m$$ is shown in the figure, each line being tangent to the envelope, for we know from Analytic Geometry that $$y = mx + \frac{p}{m}$$ is the tangent to the parabola $$y^2 = 4px$$ expressed in terms of its own slope $$m$$.

Since the normals to a curve are all tangent to the evolute, §118, p. 181, it is evident that the evolute of a curve may also be defined as the envelope of its normals; that is, as the locus of the ultimate intersections of neighboring normals. It is also interesting to notice that if we find the parametric equations of the envelope by the method of the previous section, we get the coordinates $$x$$ and $$y$$ of the center of curvature; so that we have here a second method for finding the coordinates of the center of curvature. If we then eliminate the variable parameter, we have a relation between $$x$$ and $$y$$ which is the rectangular equation of the evolute (envelope of the normals).




 * Find the evolute of the parabola $$y^2 = 4px$$ considered as the envelope of its normals.


 * Solution. The equation of the normal at any point $$(x', y')$$ is

$y - y' =-\frac{y'}{2p}(x-x')$


 * from (§2), p. 77. As we are considering the normals all along the curve, both $$x'$$ and $$y$$ will vary. Eliminating $$x'$$ by means of $$y'^2 = 4px'$$, we get the equation of the normal to be


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 * style="width:3%"|
 * align="center" |$$y - y' = \frac{y'^2}{8p^2} - \frac{xy'}{2p}$$.
 * }