Question

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 the auxillary circle at the point $M$. The area of the triangle with vertices at $A,M$ and the origin $O$ is

• A. $\frac{31}{10}$
• B. $\frac{29}{10}$
• C. $\frac{21}{10}$
• D. $\frac{27}{10}$

Answer 1

Answer: (D)

$\frac{27}{10}$

Equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{1}=1$

$\therefore a=3,b=1$

So the coordinates of $A$ are $(3,0)$ and $B$ are $(0,1)$

$\frac{x}{3}+\frac{y}{1}=1x+3y=3$ .......(i)

Equation of auxiliary circle is

$x^2+y^2=9$

Substituting $x$ from (i), we have

${(3-3y)}^2+y^2=9\Rightarrow 9+9y^2-18y+y^2=9\Rightarrow 10y^2-18y=0\Rightarrow 2y(5y-9)=0\Rightarrow y=0,\frac{9}{5}$

So, the altitude of $△$$AMO$$=y=\frac{9}{5}$

Area of $△AMO$$=\Delta =\frac{1}{2}\times a\times y$

$\Delta =\frac{1}{2}\times 3\times \frac{9}{5}=\frac{27}{10}$