The angle between the pair of tangents drawn | vedclass

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Question

(C) The equation of the ellipse is $S: 3x^2 + 2y^2 - 5 = 0$. For a point $(x_1, y_1) = (1, 2)$,the equation of the pair of tangents is given by $SS_1

Vedclass · Accepted Answer

$	an^{-1}(12/\sqrt{5})$

Vedclass · Answer

(C) The equation of the ellipse is $S: 3x^2 + 2y^2 - 5 = 0$. For a point $(x_1, y_1) = (1, 2)$,the equation of the pair of tangents is given by $SS_1 = T^2$. Here,$S_1 = 3(1)^2 + 2(2)^2 - 5 = 3 + 8 - 5 = 6$. The tangent $T$ at $(1, 2)$ is $3(1)x + 2(2)y - 5 = 3x + 4y - 5$. Thus,$(3x^2 + 2y^2 - 5)(6) = (3x + 4y - 5)^2$. Expanding this: $18x^2 + 12y^2 - 30 = 9x^2 + 16y^2 + 25 + 24xy - 30x - 40y$. Rearranging into the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$: $9x^2 - 24xy - 4y^2 + 30x + 40y - 55 = 0$. Here,$a = 9$,$h = -12$,and $b = -4$. 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} = \frac{12}{\sqrt{5}}$. Therefore,$	heta = 	an^{-1}(12/\sqrt{5})$.

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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