question_answer

The line passing through the extremity A of the major axis and the 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 A, M. and the origin O is

A) \[\frac{31}{10}\]

B) \[\frac{29}{10}\]

C) \[\frac{21}{10}\]

D) \[\frac{27}{10}\]

Correct Answer: D

Solution :

[d] Equation of the ellipse is \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{1}=1\]

An end of the major axis A be say (3, 0) and an end of the minor axis B be say (0,1). Equations of AB is therefore.\[\frac{x}{3}+\frac{y}{1}=1\]              ? (1)

Equation of the auxiliary circle is \[{{x}^{2}}+{{y}^{2}}=9\]

? (2)

Solving the equation (1) and (2) we get

\[{{x}^{2}}+{{\left( 1-\frac{x}{3} \right)}^{2}}=9\Rightarrow {{x}^{2}}+1+\frac{{{x}^{2}}}{9}-\frac{2x}{3}=9\]

\[\Rightarrow 5{{x}^{2}}-3x-36=0\Rightarrow (5x+12)(x-3)=0\]

\[\therefore x=-\frac{12}{5}\Rightarrow y=1-\frac{1}{3}\left( -\frac{12}{5} \right)=\frac{9}{5}\]

\[\therefore \] Coordinates of M are \[\left( -\frac{12}{5},\frac{9}{5} \right)\]

area of \[\Delta AOM=\frac{1}{2}.OA.MN=\frac{1}{2}\times 3\times \frac{9}{5}=\frac{27}{10}\]