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(A) The equation of the ellipse is $3x^2 + 2y^2 - 5 = 0$. Let the point be $(x_1, y_1) = (1, 2)$. The equation of the pair of tangents from $(x_1, y_1

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

Vedclass · Answer

(A) The equation of the ellipse is $3x^2 + 2y^2 - 5 = 0$. Let the point be $(x_1, y_1) = (1, 2)$. The equation of the pair of tangents from $(x_1, y_1)$ to the ellipse $S = 0$ is given by $SS_1 = T^2$. Here,$S = 3x^2 + 2y^2 - 5$,$S_1 = 3(1)^2 + 2(2)^2 - 5 = 3 + 8 - 5 = 6$,and $T = 3x(1) + 2y(2) - 5 = 3x + 4y - 5$. Substituting these into $SS_1 = T^2$: $(3x^2 + 2y^2 - 5)(6) = (3x + 4y - 5)^2$. $18x^2 + 12y^2 - 30 = 9x^2 + 16y^2 + 25 + 24xy - 30x - 40y$. $9x^2 - 4y^2 - 24xy + 30x + 40y - 55 = 0$. This is a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$ (after shifting origin). Here $a = 9$,$b = -4$,and $h = -12$. The angle $	heta$ between the 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(6\sqrt{5})}{5} = \frac{12\sqrt{5}}{5}$. Therefore,$	heta = 	an^{-1}\left(\frac{12\sqrt{5}}{5}ight)$.

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

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