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(B) The parametric equation of the line passing through $P(2,3)$ with angle $	heta = 30^{\circ}$ is given by $\frac{x-2}{\cos 30^{\circ}} = \frac{y-3

Vedclass · Accepted Answer

$17(\sqrt{3}+1)$

Vedclass · Answer

(B) The parametric equation of the line passing through $P(2,3)$ with angle $	heta = 30^{\circ}$ is given by $\frac{x-2}{\cos 30^{\circ}} = \frac{y-3}{\sin 30^{\circ}} = r$. Thus,$x = 2 + r \cos 30^{\circ} = 2 + r \frac{\sqrt{3}}{2}$ and $y = 3 + r \sin 30^{\circ} = 3 + \frac{r}{2}$. Substituting these into the equation $x^2 - 2xy - y^2 = 0$: $(2 + r \frac{\sqrt{3}}{2})^2 - 2(2 + r \frac{\sqrt{3}}{2})(3 + \frac{r}{2}) - (3 + \frac{r}{2})^2 = 0$. Expanding this,we get: $(4 + 2\sqrt{3}r + \frac{3}{4}r^2) - 2(6 + r + \frac{3\sqrt{3}}{2}r + \frac{\sqrt{3}}{4}r^2) - (9 + 3r + \frac{1}{4}r^2) = 0$. Grouping the terms of $r^2$,$r$,and the constant: $r^2(\frac{3}{4} - \frac{\sqrt{3}}{2} - \frac{1}{4}) + r(2\sqrt{3} - 2 - 3\sqrt{3} - 3) + (4 - 12 - 9) = 0$. $r^2(\frac{1}{2} - \frac{\sqrt{3}}{2}) - r(5 + \sqrt{3}) - 17 = 0$. $r^2(\frac{1-\sqrt{3}}{2}) - r(5 + \sqrt{3}) - 17 = 0$. The product of the roots $PA \cdot PB = |r_1 r_2| = |\frac{c}{a}| = |\frac{-17}{(1-\sqrt{3})/2}| = |\frac{-34}{1-\sqrt{3}}| = |\frac{34}{\sqrt{3}-1}|$. Rationalizing the denominator: $\frac{34(\sqrt{3}+1)}{3-1} = \frac{34(\sqrt{3}+1)}{2} = 17(\sqrt{3}+1)$.

$A$ line passing through $P(2,3)$ and making an angle of $30^{\circ}$ with the positive direction of the $x$-axis meets the curve $x^2-2xy-y^2=0$ at $A$ and $B$. Then the value of $PA \cdot PB$ is

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