Find the points on the curve x^{2}+y^{2}-2x-3 | vedclass

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(A) The equation of the given curve is $x^{2}+y^{2}-2x-3=0$. On differentiating with respect to $x$,we get: $2x + 2y \frac{dy}{dx} - 2 = 0$ $2y \frac{

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$(1, 2)$ and $(1, -2)$

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(A) The equation of the given curve is $x^{2}+y^{2}-2x-3=0$. On differentiating with respect to $x$,we get: $2x + 2y \frac{dy}{dx} - 2 = 0$ $2y \frac{dy}{dx} = 2 - 2x$ $\frac{dy}{dx} = \frac{1-x}{y}$. For the tangent to be parallel to the $x$-axis,the slope $\frac{dy}{dx}$ must be $0$. So,$\frac{1-x}{y} = 0$,which implies $1-x = 0$,or $x = 1$. Substituting $x = 1$ into the curve equation: $(1)^{2} + y^{2} - 2(1) - 3 = 0$ $1 + y^{2} - 2 - 3 = 0$ $y^{2} - 4 = 0$ $y^{2} = 4$ $y = \pm 2$. Thus,the points are $(1, 2)$ and $(1, -2)$.

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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