The angle between the pair of tangents drawn | vedclass

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(C) The equation of the pair of tangents from point $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $SS_1 = T^2$.For t

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$	an^{-1}\left(\frac{12}{\sqrt{5}}ight)$

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(C) The equation of the pair of tangents from point $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $SS_1 = T^2$.For the ellipse $3x^2 + 2y^2 - 5 = 0$ and point $(1, 2)$:$S = 3x^2 + 2y^2 - 5$$S_1 = 3(1)^2 + 2(2)^2 - 5 = 3 + 8 - 5 = 6$$T = 3x(1) + 2y(2) - 5 = 3x + 4y - 5$Substituting into $SS_1 = T^2$:$6(3x^2 + 2y^2 - 5) = (3x + 4y - 5)^2$$18x^2 + 12y^2 - 30 = 9x^2 + 16y^2 + 25 + 24xy - 30x - 40y$$9x^2 - 24xy - 4y^2 + 30x + 40y - 55 = 0$This is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,where $a = 9, h = -12, b = -4$.The angle $	heta$ between the pair of lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.$	an 	heta = \left| \frac{2\sqrt{(-12)^2 - (9)(-4)}}{9 - 4} ight| = \left| \frac{2\sqrt{144 + 36}}{5} ight| = \frac{2\sqrt{180}}{5} = \frac{2 	imes 6\sqrt{5}}{5} = \frac{12\sqrt{5}}{5} = \frac{12}{\sqrt{5}}$.Therefore,$	heta = 	an^{-1}\left(\frac{12}{\sqrt{5}}ight)$.

The angle between the pair of tangents drawn from the point $(1, 2)$ to the ellipse $3x^2 + 2y^2 = 5$ is:

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