Question

Find the equation of the tangents to the curve x² + y² – 2x – 4y + 1 =0 which a parallel to the X-axis.

Answer 1

Let P(x₁,y₁) be the point on the curve x² + y² – 2x – 4y + 1 = 0 where the tangent  is paralle to X-axis.   

Differentiating x² + y² – 2x – 4y + 1 = 0 w.r.t. x, we get

`2x + 2ydy/dx  2 xx 1 - 4dy/dx + 0` = 0

∴ `(2y - 4)dy/dx` = 2 – 2x

∴ `d/dx = (2 - 2x)/(2y- 4) = (1 - x)/(y - 2)`

∴ `(dy/dx)_("at" (x_1, y_1)) =(1 - x_1)/(y_1 - 2)`

= slope of the tangent at (x₁, y₁)
Since, the tangent is parallel to X-axis,
slope of the tangent  0.

∴ `(1 - x_1)/(y_1 - 2)` = 0

∴ 1 – x₁ = 0
∴ x₁ = 1
Since, (x₁, y₁) lies on x² + y² – 2x – 4y + 1 = 0,
x₁² + y₁² – 2x₁ – 4y₁ + 1 = 0
When x₁ = 1, (1)² + y₁² – 2(1) – 4y₁ + 1 = 0
∴ 1 +y₁² – 2  – 4y₁ + 1 = 0
∴ y₁² – 4y₁ = 0
∴ y₁(y₁ – 4) = 0
∴ y₁ = 0 or y₁ = 4
∴ the coordinates of the point are (1, 0) or (1, 4)
Since, the tangents are parallel to X-axis their equations are of the form y = k

If it passes through the point (1, 0), k = 0 and if it passes through the point (1, 4), k = 4

Hence, te equations of the tangents are y = 0 and y = 4.