Question

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

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

`2x + 2y(dy)/(dx) - 2 xx 1 - 4(dy)/(dx) + 0` = 0

∴ `(2y - 4)(dy)/(dx)` = 2 – 2x

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

∴ `(dy/dx)_(at(x_1,y_2)) = (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

`x_1^2 + y_1^2 - 2x_1 - 4y_1 + 1` = 0

When x₁ = `1,(1)^2 + y_1^2 - 2(1) - 4y_1 + 1` = 0

∴ 1 + y₁ – 2 – 4y₁ + 1 = 0

∴ `y_1^2 - 4y_1` = 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 it passes through the point (1, 4), k = 4

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