Question

The differential equation for all the straight lines which are at the distance of 2 units from the origin is ______.

पर्याय

• `1/4 (y - x ("d"y)/("d"x))^2 = 1 - (("d"y)/("d"x))^2`

• `1/4 (y + x ("d"y)/("d"x))^2 = 1 + (("d"y)/("d"x))^2`

• `1/4 (y - x ("d"y)/("d"x))^2 = 1 + (("d"y)/("d"x))^2`

• `1/4 (y + x ("d"y)/("d"x))^2 = 1 - (("d"y)/("d"x))^2`

Answer 1

The differential equation for all the straight lines which are at the distance of 2 units from the origin is `1/4 (y - x ("d"y)/("d"x))^2 = 1 + (("d"y)/("d"x))^2`.

Explanation:

The equation of the family of lines which are at the distance of2 units from the origin is

x cos α + y sin α = 2  .......(i)

Differentiating w.r.t. x, we get

`cos alpha + sin alpha ("d"y)/("d"x)` = 0  ......(ii)

By (i) – x × (ii), we get

`sin alpha (y - x ("d"y)/("d"x))` = 2

⇒ `y - x ("d"y)/("d"x)` = 2 cosec α  ......(iii)

From (ii), `(("d"y)/("d"x))^2 = cot^2alpha = "cosec"^2alpha - 1`

∴ `(("d"y)/("d"x))^2 = 1/4 (y - x ("d"y)/("d"x))^2 - 1`  ......[From (iii)]

∴ `1 + (("d"y)/("d"x))^2 = 1/4 (y - x ("d"y)/("d"x))^2`