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JEE Main
Exam
JEE Main
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Mathematics
Chapter
Ellipse

JEE Main · Mathematics · Ellipse

The line passing through the extremity A of the major axis and...

The line passing through the extremity AA of the major axis and extremity BB of the minor axis of the ellipse x2+9y2=9x^{2} + 9y^{2} = 9 meets its auxiliary circle at the point MM. Then, the area of the triangle with vertices at A,MA, M and the origin OO isThe line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x^ 2 + 9y^

Options

  1. A3110\dfrac{31}{10}
  2. B2910\dfrac{29}{10}
  3. C2110\dfrac{21}{10}
  4. D2710\dfrac{27}{10}

Correct answer

D. 2710\dfrac{27}{10}

Step-by-step solution

Equation of auxiliary circle is

x2+y2=9..(i)x^{2} + y^{2} = 9 . . \left(i\right)

The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x^ 2 + 9y^

∴∴ Equation of AMAM is \dfrac{x}{3} + \dfrac{y}{1} =1...(ii)1 . . . \left(ii\right)

On solving Eqs. (i) and (ii), we get

M(−125,95)</p><p>Now,areaofΔM \left(- \dfrac{12}{5}, \dfrac{9}{5}\right)</p><p>Now, area of \DeltaAOM = 12\dfrac{1}{2}OA \timesMN = 2710squnits</p><p>Hence,option(d)iscorrect.</p>\dfrac{27}{10}sq units</p><p>Hence, option (d) is correct.</p>

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