The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse $x^{2} + 9 y^{2} = 9$ meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and O(origin) is_______Sq. units.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{31}{10}$

$\frac{21}{10}$

$\frac{27}{10}$

$\frac{29}{10}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Question Image

$\begin{array}{l} \frac{x^{2}}{9} + \frac{y^{2}}{1} = 1 \\ A (3, 0), B (0, 1) \\ A B e q u a t i o n \frac{x}{3} + \frac{y}{1} = 1 \Rightarrow x + 3 y = 3 \\ d = p e r p e n d i c u l a r d i s \tan c e f r o m 0 t o A B \\ = \frac{|- 3|}{\sqrt{1 + 9}} = \frac{3}{\sqrt{10}} \\ A u x i l i a r y c i r c l e x^{2} + y^{2} = 9 \\ A M = 2 \sqrt{r^{2} - d^{2}} = 2 \sqrt{9 - \frac{9}{10}} = 2 (3) \sqrt{\frac{9}{10}} = \frac{18}{\sqrt{10}} \\ A r e a o f Δ O A M = \frac{1}{2} (\frac{18}{\sqrt{10}}) (\frac{3}{\sqrt{10}}) = \frac{27}{10} s q . u n i t s \end{array}$