The acute angle between the pair of tangents | vedclass

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(B) The equation of the ellipse is $2x^{2} + 3y^{2} = 5$,which can be written as $\frac{x^{2}}{5/2} + \frac{y^{2}}{5/3} = 1$. Here $a^{2} = \frac{5}{2

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$	an^{-1}\left(\frac{24}{7\sqrt{5}}ight)$

Vedclass · Answer

(B) The equation of the ellipse is $2x^{2} + 3y^{2} = 5$,which can be written as $\frac{x^{2}}{5/2} + \frac{y^{2}}{5/3} = 1$. Here $a^{2} = \frac{5}{2}$ and $b^{2} = \frac{5}{3}$. The equation of any tangent to the ellipse is $y = mx \pm \sqrt{a^{2}m^{2} + b^{2}} = mx \pm \sqrt{\frac{5}{2}m^{2} + \frac{5}{3}}$. Since the tangent passes through $(1, 3)$,we have $3 - m = \pm \sqrt{\frac{5}{2}m^{2} + \frac{5}{3}}$. Squaring both sides: $(3 - m)^{2} = \frac{5}{2}m^{2} + \frac{5}{3}$. $9 - 6m + m^{2} = \frac{5}{2}m^{2} + \frac{5}{3}$. Multiplying by $6$: $54 - 36m + 6m^{2} = 15m^{2} + 10$. $9m^{2} + 36m - 44 = 0$. Let $m_{1}$ and $m_{2}$ be the roots of this quadratic equation. Then $m_{1} + m_{2} = -\frac{36}{9} = -4$ and $m_{1}m_{2} = -\frac{44}{9}$. The angle $	heta$ between the tangents is given by $	an 	heta = \left|\frac{m_{1} - m_{2}}{1 + m_{1}m_{2}}ight|$. $|m_{1} - m_{2}| = \sqrt{(m_{1} + m_{2})^{2} - 4m_{1}m_{2}} = \sqrt{(-4)^{2} - 4(-\frac{44}{9})} = \sqrt{16 + \frac{176}{9}} = \sqrt{\frac{144 + 176}{9}} = \sqrt{\frac{320}{9}} = \frac{8\sqrt{5}}{3}$. $1 + m_{1}m_{2} = 1 - \frac{44}{9} = -\frac{35}{9}$. $	an 	heta = \left|\frac{8\sqrt{5}/3}{-35/9}ight| = \left|\frac{8\sqrt{5}}{3} 	imes \frac{-9}{35}ight| = \frac{24\sqrt{5}}{35} = \frac{24\sqrt{5}}{7 	imes 5} = \frac{24}{7\sqrt{5}}$. Thus,$	heta = 	an^{-1}\left(\frac{24}{7\sqrt{5}}ight)$.

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

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