The line passing through the extremity of the major axis and extremity of the minor axis of the ellipse meets its auxiliary circle at the point . Then the area of the triangle with vertices at and the origin is [2009]

Question

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 the origin $O$ is [2009]

Smart Achievers · Accepted Answer

(4)

$Given ellipse is x^{2} + 9 y^{2} = 9 \Rightarrow \frac{x^{2}}{3^{2}} + \frac{y^{2}}{1^{2}} = 1$

$∴$ $Coordinates of A and B are (3, 0) and (0, 1) respectively$

$∴$ $Equation of A B is \frac{x}{3} + \frac{y}{1} = 1$

$\Rightarrow x + 3 y - 3 = 0 . . . (i)$

$and equation of auxiliary circle of given ellipse is$

$x^{2} + y^{2} = 9 . . . (i i)$

$On solving equations (i) and (ii), we get the point M where line A B meets the auxiliary circle.$

$Putting x = 3 - 3 y from (i) in (ii), we get$

${(3 - 3 y)}^{2} + y^{2} = 9$

$\Rightarrow 9 - 18 y + 9 y^{2} + y^{2} = 9$ $\Rightarrow 10 y^{2} - 18 y = 0$

$\Rightarrow y = 0, \frac{9}{5} \Rightarrow x = 3, \frac{- 12}{5}$

$Clearly M (\frac{- 12}{5}, \frac{9}{5})$

$∴$ $Area of ∆ O A M = \frac{1}{2}$ $| \begin{matrix} 0 & 0 & 1 \\ 3 & 0 & 1 \\ \frac{- 12}{5} & \frac{9}{5} & 1 \end{matrix} |$ $= \frac{27}{10}$

Solution

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 the origin $O$ is [2009]

1 $\frac{31}{10}$

2 $\frac{29}{10}$

3 $\frac{21}{10}$

4 $\frac{27}{10}$

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Solution

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 the origin O is [2009]

1 3110

2 2910

3 2110