Question

The normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.

Answer 1

Let P(x, y) be a point on the curve y = f(x).

Then slope of the normal to the curve is `- 1/("dy"/"dx")`

∴ equation of the normal is

`("Y - y") = - 1/("dy"/"dx") ("X - x")`

∴ `("Y - y") "dy"/"dx" = - ("X - x")`

∴ `("Y - y") "dy"/"dx" + ("X - x") = 0`

Since, this normal passes through (2, 0), we get

`(0 - "y") "dy"/"dx" + (2 - "x") = 0`

∴ `- "y" "dy"/"dx" = "x - 2"`

∴ `- "y"  "dy" = ("x - 2")"dx"`

Integrating both sides, we get

`int - "y"  "dy" = int ("x - 2")"dx"`

∴ `- "y"^2/2 = "x"^2/2 - 2"x" + "c"_1`

∴ `"x"^2/2 + "y"^2/2 - 2"x" + "c"_1 = 0`

∴ x² + y² - 4x + 2c₁ = 0 

∴ x² + y² = 4x - 2c₁

∴ x² + y² = 4x + c, where  c = - 2c₁

This is the general equation of the curve.

Since, the required curve passed through the point (2, 3), we get 

2² + 3² = 4(2) + c

∴ c = 5

∴ equation of the required curve is

x² + y² = 4x + 5.