• question_answer If $(\sqrt{3})bx+ay=2ab$touches the ellipse $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1,$then the eccentric angle $\theta$ of the point of contact is A)  $\pi /6$                             B)  $\pi /2$ C)  $3\pi /4$                          D)  $2\pi /3$

Idea Equation of tangent of an ellipse $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ at$\theta$i.e., $(a\cos \theta ,b\sin \theta )$is$\frac{x}{a}\cos \theta +\frac{y}{b}\cos \theta =1$Now, compare the both equations of tangent to get $\theta$. Here, equation of tangent $\frac{x}{a}\frac{\sqrt{3}}{2}+\frac{y}{b}\frac{1}{2}=1$ and equation of tangent at the point (a$\cos \theta ,b\sin \theta$) is$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ Both are same i. e., $\cos \theta +\frac{\sqrt{3}}{2},\sin \theta =\frac{1}{2}\Rightarrow$$\theta =\frac{\pi }{6}$ TEST Edge Equation of normal, latus rectum, based questions are asked. To solve these types of questions, students are advised to understand the concept of an ellipse.
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