If the line y - sqrt{3}x + 3 = 0 cuts the par | vedclass

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Question

(A) The line equation is $y - \sqrt{3}x + 3 = 0$. We can write it in parametric form as $\frac{x - \sqrt{3}}{\cos 	heta} = \frac{y - 0}{\sin 	heta}

Vedclass · Accepted Answer

$\frac{4(\sqrt{3} + 2)}{3}$

Vedclass · Answer

(A) The line equation is $y - \sqrt{3}x + 3 = 0$. We can write it in parametric form as $\frac{x - \sqrt{3}}{\cos 	heta} = \frac{y - 0}{\sin 	heta} = r$. Comparing with the line equation,the slope is $	an 	heta = \sqrt{3}$,so $	heta = 60^\circ$. Thus,$\cos 	heta = \frac{1}{2}$ and $\sin 	heta = \frac{\sqrt{3}}{2}$. The parametric coordinates are $x = \sqrt{3} + \frac{r}{2}$ and $y = \frac{\sqrt{3}r}{2}$. Substituting these into the parabola equation $y^2 = -x - 2$: $(\frac{\sqrt{3}r}{2})^2 = -(\sqrt{3} + \frac{r}{2}) - 2$ $\frac{3r^2}{4} = -\sqrt{3} - \frac{r}{2} - 2$ $3r^2 + 2r + 4(\sqrt{3} + 2) = 0$. Since $PA \cdot PB = |r_1 r_2|$,by the product of roots formula,$|r_1 r_2| = |\frac{c}{a}| = |\frac{4(\sqrt{3} + 2)}{3}| = \frac{4(\sqrt{3} + 2)}{3}$.

If the line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^2 = -x - 2$ at $A$ and $B$,then $PA \cdot PB$ is equal to,where $P \equiv (\sqrt{3}, 0)$.

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