Answer to Question #145434 in Differential Geometry | Topology for Dolly

Question

Find envelope of the family of straight line
y=mx+a/m

Answer 1

Envelope of one-parameter family of curve is the locus of the limiting positions of the points of intersection of any two members of the family when one of the family tends to coincide with the other family which is kept fixed. It is done by Eliminating '$m$' which is assumed to be parameter here from given and partial differentiation equations.

We have, $y=mx+\frac{a}{m}$ (*)

Partial differentiate with respect to $m$: $0=x+a(-\frac{1}{m^2})$. It follows that $m=\sqrt{\frac{a}{x}}$ or $m=-\sqrt{\frac{a}{x}}$ .

Substituting in equation (*) , $y=\sqrt{ax}+\sqrt{ax}=2\sqrt{ax}$ . When we take negative value of $m$, then $y=-2\sqrt{ax}$. Combining these two, we get $y^2=4ax.$

This represents Parabola which is required envelope of family of straight lines $y=mx+\frac{a}{m}$.