# At what points on the curve $x^{2}+y^{2}-2x-4y+1=0$ the tangent is parallel to $y$- axis.

Note: This is part 2nd of a 2 part question, split as 2 separate questions here.

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) Step 1: Let (x_1,y_1) be the point at which the tangent is || to the y-axis Therefore \large\frac{dy}{dx_{(x_1,y_1)}}$$=m \to \infty$
Step 2:
Now $\large\frac{dy}{dx_{(x_1,y_1)}}=\large\frac{1-x_1}{y_1-2}$$\qquad m \to \infty =>y_1-2=\infty$
=>$y=2$
Step 3:
Substituting $y_1=2$ we find $x_1$
$x_1^2+4-2x_1-8+1=0$
=>$x_1^2-2x_1-3=0$
=>$(x_1-3)(x_1+1)=0$
=>$x_1=3,-1$
Step 4:
The points are $(3,2),(-1,2)$