• # question_answer If letup rectum of the ellipse x2 tan2 $\alpha$ + y2 sec2 $\alpha$ = 1 is 1/2, then $\alpha$ $(0<\alpha <\pi /3)$ is equal to A)  $\pi /6$                             B)  $\pi /12$ C)  $\pi /4$                             D)  $3\pi /4$

Solution :

Idea An ellipse $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1,$ where ${{b}^{2}}={{a}^{2}}$$(1-{{e}^{2}})$ length of latus rectum is $\frac{2{{b}^{2}}}{a}$ Here, the ellipse is x2 tan2 a + y2 sec2 a =1 So, latus rectum $=\frac{2{{\cos }^{2}}\alpha }{\cot \alpha }=\frac{1}{2}$ So,          $\sin 2\alpha =\frac{1}{2}or2\alpha =\frac{\pi }{6},\frac{5\pi }{6}...$ TEST Edge Position of a point relative to an ellipse directories of an ellipse based question are asked. To solve these types of questions, students are advised to learn the different equations of the part of an ellipse.

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