IIT-JEE 2009 · Paper 1 · Mathematics Q25

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MathematicsSingle correct+3 / -1Question 25 of 60

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

\frac{31}{10}

\frac{29}{10}

\frac{21}{10}

\frac{27}{10}

Answer and worked solution

Official answer: D

Points. $\dfrac{x^2}9+y^2=1$ : $A=(3,0)$ , $B=(0,1)$ ; line AB is $x+3y=3$ . Auxiliary circle $x^2+y^2=9$ .

M. Substituting $x=3-3y$ : $9-18y+10y^2=9$ , so $y=\frac95$ and $x=-\frac{12}5$ .

Area of OAM. Base $OA=3$ on the x-axis, height $\frac95$ : $\frac12\cdot3\cdot\frac95=\dfrac{27}{10}$ .

(D)

Topics:Ellipse Equation of a circle

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IIT-JEE 2009 · Paper 1, Mathematics question 25