The angle between the pair of tangents drawn | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The equation of the ellipse is $3x^2 + 2y^2 - 5 = 0$. Let the point be $(x_1, y_1) = (1, 2)$. The combined 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 $T = 3x(1) + 2y(2) - 5 = 3x + 4y - 5$. So,$6(3x^2 + 2y^2 - 5) = (3x + 4y - 5)^2$. Expanding this,$18x^2 + 12y^2 - 30 = 9x^2 + 16y^2 + 25 + 24xy - 30x - 40y$. Rearranging into the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $9x^2 - 24xy - 4y^2 + 30x + 40y - 55 = 0$. Here,$a = 9$,$b = -4$,and $h = -12$. The angle $	heta$ between the pair of tangents is given by $	an 	heta = \frac{2\sqrt{h^2 - ab}}{a + b}$. $	an 	heta = \frac{2\sqrt{(-12)^2 - (9)(-4)}}{9 - 4} = \frac{2\sqrt{144 + 36}}{5} = \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 to the ellipse $3x^2 + 2y^2 = 5$ from the point $(1, 2)$ is:

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