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

Note: This is part 1st of a 2 part question, split as 2 separate questions here.

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1 Answer

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Toolbox:
  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
$x^2+y^2-2x-4y+1=0$
Step 1:
Let $(x_1,y_1)$ be the point on the curve at which the tangent is $||$ to $x-axes =>m=0$
Therefore $\large\frac{dy}{dx_(x_1,y_1)}$$=0$
Step 2:
Differentiating $x^2+y^2-2x-4y+1=0$ with respect to x
$2x+2y \large\frac{dy}{dx}$$-2-4 \large\frac{dy}{dx}=0$
$\large\frac{dy}{dx}$$(2y-4)=2-2x$
$\large\frac{dy}{dx}=\frac{1-x}{y_1-2}$
$m=\large\frac{dy}{dx_{(x_1y_1)}}=\frac{1-x_1}{y_1-2}$
Step 3:
$1+y_1^2-2-4y_1+1=0$
$y_1^2-4y_1=0=>y_1(y_1-4)=0=>y_1=0,4$
The points are $(1,0) (1,4)$
answered Jul 24, 2013 by meena.p
 

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