The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x2+9 y2= 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M, and O (the origin) is

Question

The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x2+9 y2= 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M, and O (the origin) is (A) 31 / 10 (B) 29/10 (C) 21/10 (D) 27 / 10

Tardigrade Admin · Accepted Answer

The equation of line A M is x+3 y-3=0 Perpendicular distance of line from the origin =(3/√10) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/5a216a85bc084574802e2028eeb470c1-.png" /> Length of A M=2 √9-(9/10)=2 × (9/√10) or Area =(1/2) × 2 × (9/√10) × (3/√10)=(27/10) sq. units.

Q. 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 $O$ (the origin) is

$31/10$

$29/10$

$21/10$

$27/10$

The equation of line $A M$ is
$x + 3 y - 3 = 0$
Perpendicular distance of line from the origin $= \frac{3}{10}$

Length of $A M = 2 9 - \frac{9}{10} = 2 \times \frac{9}{10}$
or Area $= \frac{1}{2} \times 2 \times \frac{9}{10} \times \frac{3}{10} = \frac{27}{10}$ sq. units.

Q. The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x2+9y2=9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A,M, and O (the origin) is

31/10

29/10

21/10

27/10