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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
+1 vote

Find the points on the curve \(x^2 + y^2 - 2x - 3 = 0\) at which the tangents are parallel to the \(x\) - axis.

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1 Answer

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Given curve is $x^2+y^2-2x-3=0$----(1)
Differentiating w.r.t $x$ we get,
$2x+2y\large\frac{dy}{dx}$$-2=0$
$\Rightarrow 2y\large\frac{dy}{dx}$$=2-2x$
$\Rightarrow \large\frac{dy}{dx}=\frac{2-2x}{2y}$
$\qquad\quad=\large\frac{1-x}{y}$
Step 2:
The tangent is parallel to $x$-axis,if $\large\frac{dy}{dx}$$=0$
(i.e)$\large\frac{1-x}{y}$$=0$
$\Rightarrow 1-x=0$
Therefore $x=1$
Step 3:
Put $x=1$ in equ(1)
$1+y^2-2-3=0$
$\Rightarrow y^2-4=0$
$y^2=4$
$\Rightarrow y=\pm 2$
Hence the required points are $(1,2)$ and $(1,-2)$
(i.e) $(1,\pm 2)$
answered Jul 11, 2013 by sreemathi.v
 

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