Question

The differential equation (1 + y²)x dx – (1 + x²)y dy = 0 represents a family of:

Options

• ellipses of constant eccentricity

• ellipses of variable eccentricity

• hyperbolas of constant eccentricity

• hyperbolas of variable eccentricity

Answer 1

hyperbolas of variable eccentricity

Explanation:

Given `(x  dx)/(1 + x^2) = (y  dy)/(1 + y^2)`

Integrating we get, `1/2 log(1 + x^2) = 1/2 log(1 + y^2) + a`

⇒ `1 + x^2 = c(1 + y^2)`, where c = e^2a

`x^2 - cy^2 = c` = 1

⇒ `x^2/(c - 1) - y^2/(((c - 1)/c))` = 1  ......(i)

Clearly c > 0 as c = e^2a

Hence, the equation (i) gives a family of hyperbolas with eccentricity = `sqrt((c - 1 + (c - 1)/c)/(c - 1)) = sqrt((c^2 - 1)/(c - 1)) = sqrt(c + 1)` if c ≠ 1.

Thus eccentricity varies from member to member of the family as it depends on c.