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  •  If the tangent at a point on the ellipse <img alt="\frac{x^{2}}{27}+\frac{y^{2}}{3}= 1" src="https://learn.careers360.com/latex-image/?%5Cdpi%7B100%7D%20%5Cfrac%7Bx%5E%7B2%7D%7D%7B27%7D&p

 If the tangent at a point on the ellipse \frac{x^{2}}{27}+\frac{y^{2}}{3}= 1meets the coordinate axes at A and B, and O is the origin, then the minimum area (in sq. units) of the triangle OAB is :

 

Option: 1

\frac{9}{2}


Option: 2

3\sqrt{3}


Option: 3

9\sqrt{3}


Option: 4

9


Answers (1)

best_answer

Equation of a tangent to the ellipse is 

\frac{x}{\sqrt{27}}\cos \Theta +\frac{y}{\sqrt{3}}\sin \Theta =1

For point A, x=0

\\\frac{0}{\sqrt{27}}\cos \Theta +\frac{y}{\sqrt{3}}\sin \Theta =1\\ \\\frac{y}{\sqrt{3}}\sin \Theta =1\\ y=\frac{\sqrt{3}}{\sin \Theta}

For point B, y=0

\\\frac{x}{\sqrt{27}}\cos \Theta +\frac{0}{\sqrt{3}}\sin \Theta =1\\ \\\frac{x}{\sqrt{27}}\cos \Theta =1\\ x=\frac{\sqrt{27}}{\cos \Theta}

Area bounded =\frac{1}{2} \:\frac{\sqrt{27}}{\cos \Theta } .\frac{\sqrt{3}}{\sin \Theta }

= \frac{9}{\sin 2\Theta }

when \sin 2\Theta =1;

we get Area = 9

Posted by

Ajit Kumar Dubey

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