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

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• If the tangent of a curve is parallel to $y$ - axis then $\large\frac{dx}{dy}=0$
Step 1
Equation of the curve is $x^2+y^2-2x-4y+1=0\: \: (1)$
differentiating w.r.t $x$ we get,
$2x+2y\large\frac{dy}{dx}-2-4\large\frac{dy}{dx}=0$
$\Rightarrow \large\frac{dy}{dx}(2y-4)=2-2x$
$\therefore \large\frac{dy}{dx}=\large\frac{(1-x)}{(y-2)}$
It is given that tangent is parallel to $y$ - axis then $\large\frac{dx}{dy}=0$
$\Rightarrow \large\frac{y-2}{1-x}=0$
$\therefore y-2=0$
$\Rightarrow y=2$
Substituting for $y=2$ in equation (1)
$x^2+4-2x-8+1=0$
$\Rightarrow x^2-2x-3=0$
$\Rightarrow (x-3)(x+1)=0$
$\therefore x=3\: and\: x=-1$
Hence the points are $(3,2)\: and\: (3, -1)$
answered Aug 4, 2013