At what points on the curve

Solution:

Given, the equation of the curve is x2 + y2 – 2x – 4y + 1 = 0 ….. (i)

Differentiating both the sides w.r.t. x, we get

$2 x+2 y \cdot \frac{d y}{d x}-2-4 \cdot \frac{d y}{d x}=0$

$\Rightarrow(2 y-4) \frac{d y}{d x}=2-2 x \Rightarrow \frac{d y}{d x}=\frac{2-2 x}{2 y-4}$ ...........(ii)

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

$\therefore$ Slope $\frac{d y}{d x}=\tan \frac{\pi}{2}=\infty=\frac{1}{0}$

So, from eq. (ii) we get

$\frac{2-2 x}{2 y-4}=\frac{1}{0} \Rightarrow 2 y-4=0 \quad \Rightarrow \quad y=2$

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

 

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