Question

find the envelope of family of straight line Y=mx+a/m

Answer 1

How to get points???

Answer 2

Answer:

The envelope of family of straight line is $y^2=4ax$

Step-by-step explanation:

Given: Line is $Y=mx+\frac{a}{m}$

To find : Find the envelope of family of straight line.

Envelope of one-parameter family of curve is defined as 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.

So, First we are going to eliminate 'm' which is assumed to be parameter here from given and partial differentiation equations.

We have,

$Y=mx+\frac{a}{m}$  .....[1]

Partial differentiate with respect to m,

$0=x+a\times\frac{-1}{m^2}$

$m=\pm\sqrt\frac{a}{x}$

Then substituting in equation [1]

$y=\sqrt{ax}+\sqrt{ax}$

$y=2\sqrt{ax}$

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.