Question

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

Answer 1

Final Answer :
${y}^{2} = 4ax$

Steps:
1) Theoretically ,
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.

2) It is done by Eliminating 'm' which is assumed to be parameter here from given and partial differentiation equations.

We have,
$y = mx + \frac{a}{m} \: \: \: - - - (1)$
Partial differentiate with respect to m,

$0 = x + a \times ( \frac{ - 1}{ {m}^{2} } ) \\ = > m \: = \sqrt{ \frac{a}{x} } \: \: \: or \: - \sqrt{ \frac{a}{x} }$

3) Substituting in equation (1) , ( any one)
$y = \sqrt{ax} + \sqrt{ax} \\ = > y = 2 \sqrt{ax}$

When we take negative value of m, then
$y = - 2 \sqrt{ax}$

4) Combining these two, we get
${y}^{2} = 4ax$
This represents Parabola (Here 'a'is parameter) which is required envelope of family of straight lines :
y=mx +a/m where m is parameter.

Answer 2

Final Answer:

y2 = 4ax

Steps:

1) Theoretically,

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.

2) It is done by Eliminating 'm' which is assumed to be parameter here from given and partial differentiation equations.

We have,

m -- (1) Partial differentiate with respect to

y = mx + a

m,

0 = x + a x () m2

=> m = a or

3) Substituting in equation (1), (any

one)

Y = Vax + vax

=> y = 2 Vax

When we take negative value of m,

then

y = -2Vax

4) Combining these two, we get

y2 = 4ax

This represents Parabola (Here 'a'is parameter) which is required envelope of family of straight lines : y=mx +a/m where m is parameter.