Conic-Sections Question 548
Question: The line passing through the extremity A of the major axis and the 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 A, M. and the origin O is
Options:
A) $ \frac{31}{10} $
B) $ \frac{29}{10} $
C) $ \frac{21}{10} $
D) $ \frac{27}{10} $
Show Answer
Answer:
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}+{{( 1-\frac{x}{3} )}^{2}}=9\Rightarrow x^{2}+1+\frac{x^{2}}{9}-\frac{2x}{3}=9 $
$ \Rightarrow 5x^{2}-3x-36=0\Rightarrow (5x+12)(x-3)=0 $
$ \therefore x=-\frac{12}{5}\Rightarrow y=1-\frac{1}{3}( -\frac{12}{5} )=\frac{9}{5} $
$ \therefore $ Coordinates of M are $ ( -\frac{12}{5},\frac{9}{5} ) $ area of $ \Delta AOM=\frac{1}{2}.OA.MN=\frac{1}{2}\times 3\times \frac{9}{5}=\frac{27}{10} $
BETA
