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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.

1 Answer

  • 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$
Step 3:
Substituting $y_1=2$ we find $x_1$
=>$ (x_1-3)(x_1+1)=0$
Step 4:
The points are $ (3,2),(-1,2)$
answered Jul 24, 2013 by meena.p

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