Question

Find the equation of a curve passing through the point (1, 1) if the perpendicular distance of the origin from the normal at any point P(x, y) of the curve is equal to the distance of P from the x-axis.

Answer 1

Let the equation of normal at P(x, y) be Y – y = `(-"dx")/"dy" ("X" - x)`

i.e., `"Y" + "X" "dx"/"dy" - (y + x "dx"/"dy")` = 0   .....(1)

Therefore, the length of perpendicular from origin to (1) is

`(y + x "dx"/"dy")/sqrt(1 + ("dx"/"dy")^2)`  .....(2)

Also distance between P and x-axis is |y|.

Thus, we get `(y + x "dx"/"dy")/sqrt(1 + ("dx"/"dy")^2) = |y|`

⇒ `(y + x "dx"/"dy")^2 = y^2 [1 + ("dx"/"dy")^2]`

⇒ `"dx"/"dy" ["dx"/"dy" (x^2 - y^2) + 2xy]` = 0

⇒ `"dx"/"dy"` = 0

or `"dx"/"dy" = (2xy)/(y^2 - x^2)`

Case I: `"dx"/"dy" = 0

⇒ dx = 0

Integrating both sides, we get x = k,

Substituting x = 1, we get k = 1.

Therefore, x = 1 is the equation of curve  .....(not possible, so rejected).

Case II: `"dx"/"dy" = (2xy)/(y^2 - x^2)`

⇒ `"dy"/"dx" = (y^2 - x^2)/(2xy)`.

Substituting y = vx, we get

`"v" + x "dv"/"dx" = ("v"^2x^2 - x^2)/(2"v"x^2)`

⇒ `x * "dv"/"dx" = ("v"^2 - 1)/(2"v")`

= `(-(1 + "v"^2))/(2"v")`

⇒ `(2"v")/(1 + "v"^2) "dv" = (-"dv")/x`

Integrating both sides, we get

log(1 + v²) = – logx + logc

⇒ log(1 + v²)(x) = log c

⇒ (1 + v²) x = c

⇒ x² + y² = cx.

Substituting x = 1,

y = 1, we get c = 2.

Therefore, x² + y² – 2x = 0 is the required equation.