For the two curves C_1 : y^2 = 4x and C_2 : x | vedclass

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(C) Substituting $y^2 = 4x$ into the equation of $C_2$:$x^2 + 4x - 6x + 1 = 0$$\Rightarrow x^2 - 2x + 1 = 0$$\Rightarrow (x - 1)^2 = 0$$\Rightarrow x

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$C_1$ and $C_2$ touch each other at exactly two points.

Vedclass · Answer

(C) Substituting $y^2 = 4x$ into the equation of $C_2$:$x^2 + 4x - 6x + 1 = 0$$\Rightarrow x^2 - 2x + 1 = 0$$\Rightarrow (x - 1)^2 = 0$$\Rightarrow x = 1$.For $x = 1$,$y^2 = 4(1) = 4$,so $y = \pm 2$.The points of intersection are $(1, 2)$ and $(1, -2)$.For $C_1$,the derivative is $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$. At $(1, 2)$,slope $m_1 = 1$. At $(1, -2)$,slope $m_1 = -1$.For $C_2$,$2x + 2y \frac{dy}{dx} - 6 = 0 \Rightarrow \frac{dy}{dx} = \frac{3-x}{y}$. At $(1, 2)$,slope $m_2 = \frac{3-1}{2} = 1$. At $(1, -2)$,slope $m_2 = \frac{3-1}{-2} = -1$.Since the slopes are equal at both points,the curves touch each other at exactly two points.

For the two curves $C_1 : y^2 = 4x$ and $C_2 : x^2 + y^2 - 6x + 1 = 0$,which of the following is true?

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