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

Given: equation of a curve x² + y² – 2x – 4y + 1 = 0

To find: the points on the curve x² + y² – 2x – 4y + 1 = 0, the tangents are parallel to the y-axis

Explanation:

Now given equation of curve as x² + y² – 2x – 4y + 1 = 0

Differentiating this with respect to x, we get

Applying the sum rule of differentiation, we get

As it is given the tangents are parallel to the y-axis,

Hence

Substituting the value from equation (i), we get

⇒ y-2 = 0

⇒ y = 2

Now substituting y = 2 in curve equation, we get

x² + y² – 2x – 4y + 1 = 0

⇒ x² + 2² – 2x – 4(2) + 1 = 0

⇒ x² + 4 – 2x – 8 + 1 = 0

⇒ x²– 2x-3 = 0

Now splitting the middle term, we get

⇒ x²-3x+x-3 = 0

⇒ x(x-3)+1(x-3) = 0

⇒ (x+1)(x-3) = 0

⇒ x+1 = 0 or x-3 = 0

⇒ x = -1 or x = 3

So the required points are (-1, 2) and (3, 2).

Hence the points on the curve x² + y² – 2x – 4y + 1 = 0, the tangents are parallel to the y-axis are (-1, 2) and (3, 2).