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What are the points on the curve $x^{2}+y^{2}-2 x-3=0$ where the tangents are parallel tox-axis?
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The correct answer is:
$(1,2)$ and $(1,-2)$
Given equation curve is $x^{2}+y^{2}-2 x-3=0$
On differentiating we get
$2 x+2 y \frac{d y}{d x}-2=0$
$\Rightarrow y \frac{d y}{d x}=-x+1$
$\Rightarrow \frac{d y}{d x}=-\frac{x+1}{y}$
Since, tangent is parallel to $x$ -axis
$\therefore \frac{d y}{d x}=0 \Rightarrow-\frac{x+1}{y}=0 \Rightarrow x=1$
From equation (1), we have $1+y^{2}-2-3=0 \Rightarrow y=\pm 2$
Hence, required points are $(1,2)$ and $(1,-2)$.
On differentiating we get
$2 x+2 y \frac{d y}{d x}-2=0$
$\Rightarrow y \frac{d y}{d x}=-x+1$
$\Rightarrow \frac{d y}{d x}=-\frac{x+1}{y}$
Since, tangent is parallel to $x$ -axis
$\therefore \frac{d y}{d x}=0 \Rightarrow-\frac{x+1}{y}=0 \Rightarrow x=1$
From equation (1), we have $1+y^{2}-2-3=0 \Rightarrow y=\pm 2$
Hence, required points are $(1,2)$ and $(1,-2)$.
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