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  • If the tangent at the point <img alt="\mathrm{\left(4 \cos \theta, \frac{16}{\sqrt{11}} \sin \theta\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%284%20%5Ccos%20

If the tangent at the point \mathrm{\left(4 \cos \theta, \frac{16}{\sqrt{11}} \sin \theta\right)} to the ellipse \mathrm{16 x^2+11 y^2=256} is also a tangent to the circle \mathrm{x^2+y^2-2 x=15}, then \mathrm{\theta} equals

Option: 1

\frac{\pi}{3}


Option: 2

\frac{2\pi}{3}


Option: 3

-\frac{\pi}{3}


Option: 4

 Both (a) and (c) 


Answers (1)

best_answer

\mathrm{\text { Equation of tangent to } 16 x^2+11 y^2=256 \text { at }}

\mathrm{\begin{aligned} & \left(4 \cos \theta, \frac{16}{\sqrt{11}} \sin \theta\right) \text { is } \\ & 4 \cos \theta x+\sqrt{11} \sin \theta y=16 \end{aligned}}

The perpendicular from centre (1, 0) is equal to radius

\mathrm{\begin{aligned} & \sqrt{(1+15)}=4 \\ & \text { or } \frac{|4 \cos \theta-16|}{\sqrt{\left(16 \cos ^2 \theta+11 \sin ^2 \theta\right)}}=4 \end{aligned}}

On simplification, we get

\mathrm{\cos \theta=\frac{1}{2} \text { or }-\frac{5}{2} \text { (not possible) }}

\mathrm{\text { or } \theta= \pm \frac{\pi}{3}, \frac{5 \pi}{3}}

Posted by

Ritika Kankaria

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