Question

At what points on the curve x² + y² – 2x – 4y + 1 = 0, the tangents are parallel to the y-axis?

Answer 1

Given that the equation of the curve is

x² + y² – 2x – 4y + 1 = 0  ....(i)

Differentiating both sides w.r.t. x, we have

`2x + 2y * "dy"/"dx" - 2 - 4 * "dy"/"dx"` = 0

⇒ `(2y - 4) "dy"/"dx"` = 2 – 2x

⇒ `"dy"/"dx" = (2 - 2x)/(2y - 4)`  ....(ii)

Since the tangent to the curve is parallel to the y-axis.

∴ Slope `"dy"/"dx" = tan  pi/2`

= `oo`

= `1/0`

So, from equation (ii) we get

`(2 - 2x)/(2y - 4) = 1/0`

⇒ 2y – 4 = 0

⇒ y = 2

Now putting the value of y in equation (i), we get

⇒ x² + (2)² – 2x – 8 + 1 = 0

⇒ x² – 2x + 4 – 8 + 1 = 0

⇒ x² – 2x – 3 = 0

⇒ x² – 3x + x – 3 = 0

⇒ x(x – 3) + 1(x – 3) = 0

⇒ (x – 3)(x + 1) = 0

⇒ x = – 1 or 3

Hence, the required points are (– 1, 2) and (3, 2).