# 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).

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