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Home » Find the maximum area of an isosceles triangle inscribed in the ellipse x^2/a^2+^2/b^2 with its vertex at one end of the major axis.
Find the maximum area of an isosceles triangle inscribed in the ellipse x^2/a^2+^2/b^2 with its vertex at one end of the major axis.
Applications of Derivatives | CBSE | Class 12 | Maths | Miscellaneous | NCERT
Find the maximum area of an isosceles triangle inscribed in the ellipse x^2/a^2+^2/b^2 with its vertex at one end of the major axis.

 Equation of the ellipse is  ……….(i)

Comparing eq. (i) with  we have  and 

 and 

Any point on ellipse is P

Draw PM perpendicular to axis and produce it to meet the ellipse in the point Q.

 OM =  and PM = 

We know that the ellipse (i) is symmetrical about axis, therefore, PM = QM and hence triangle APQ is isosceles.

Area of APQ  x Base x Height

=  PQ.AM =  . 2PM.AM = PM (OA – OM)

= 

= 

 

= 

 

Now 

 = 0

 

 

 

  or 

i.e.,  or 

 

 is impossible

At , 

=  [Negative]

 is maximum at 

From eq. (i), Maximum area

= 

= 

=  = 

i

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