JEE Advanced · Mathematics · 14. Ellipse
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
- A
\(\frac{31}{10}\)
- B
\(\frac{29}{10}\)
- C
\(\frac{21}{10}\)
- D
\(\frac{27}{10}\)
Answer & Solution
Correct Answer
(D)
\(\frac{27}{10}\)
Step-by-step Solution
Detailed explanation
Equation of auxiliary circle is
\[
x^2+y^2=9
\]
Equation of \(A M\) is \(\frac{x}{3}+\frac{y}{1}=1\)

On solving Eqs. (i) and (ii), we get Coordinates of \(M=\left(-\frac{12}{5}, \frac{9}{5}\right)\) Now, area of \(\triangle A O M=\frac{1}{2} \times O A \times M N\) \(=\frac{27}{10}\) sq units
\[
x^2+y^2=9
\]
Equation of \(A M\) is \(\frac{x}{3}+\frac{y}{1}=1\)

On solving Eqs. (i) and (ii), we get Coordinates of \(M=\left(-\frac{12}{5}, \frac{9}{5}\right)\) Now, area of \(\triangle A O M=\frac{1}{2} \times O A \times M N\) \(=\frac{27}{10}\) sq units
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