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The point on the curve $x^2+y^2=a^2, y \geq 0$, at which the tangent is parallel to the $x$-axis is
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Verified Answer
The correct answer is:
$(0, a)$
We have,
$x^2+y^2=a^2$
$\Rightarrow \quad 2 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{y}$
Tangent is parallel to $X$-axis.
$\therefore \quad \frac{d y}{d x}=0$
$\Rightarrow \quad \frac{-x}{y}=0 ;
\therefore x=0$
Putting the value of $x$ in $x^2+y^2=a^2$
$\Rightarrow \quad 0+y^2=a^2 \Rightarrow y=a, y \geq 0$
$\therefore$ The point on the curve is $(0, a)$.
$x^2+y^2=a^2$
$\Rightarrow \quad 2 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-x}{y}$
Tangent is parallel to $X$-axis.
$\therefore \quad \frac{d y}{d x}=0$
$\Rightarrow \quad \frac{-x}{y}=0 ;
\therefore x=0$
Putting the value of $x$ in $x^2+y^2=a^2$
$\Rightarrow \quad 0+y^2=a^2 \Rightarrow y=a, y \geq 0$
$\therefore$ The point on the curve is $(0, a)$.
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