Question

At what point on the curve $x^2+y^2-2x-4y+1=0$, the tangents are parallel to the y axis

Answer 1

$x^2+y^2-2x-4y+1$

$\Rightarrow 2x+2y\frac{dy}{dx}-2-4\frac{dy}{dx}=0\Rightarrow x+y\frac{dy}{dx}-1-2\frac{dy}{dx}=0\Rightarrow (y-2)\frac{dy}{dx}=(1-x)\frac{dy}{dx}=\frac{(y-2)}{(1-x)}$

for the tangents to be parallel to $y-$ axis, $\frac{dy}{dx}=0$

$\therefore \frac{dy}{dx}=\frac{(y-2)}{(1-x)}=0\Rightarrow y=2$

When $y=2$

$x^2+2^2-2x-4\left(2\right)+1=0\Rightarrow x^2+4-2x-8+1=0\Rightarrow x^2-2x-3=0\Rightarrow (x-1)(x-3)=0\Rightarrow x=-1or3$

So, the points where tangents are parallel to $y-$ axis

$=(-1,2),(3,2)$