Menu Close

Question-47519




Question Number 47519 by ajfour last updated on 11/Nov/18
Commented by ajfour last updated on 11/Nov/18
Find minimum area of △ABC  in terms of ellipse parameters  a and b.
$${Find}\:{minimum}\:{area}\:{of}\:\bigtriangleup{ABC} \\ $$$${in}\:{terms}\:{of}\:{ellipse}\:{parameters} \\ $$$${a}\:{and}\:{b}. \\ $$
Answered by ajfour last updated on 11/Nov/18
Minimum Area(△ABC )= △_(min)   △_(min) =(((1+m^2 )^2 )/(m((m^2 /a^2 )+(1/b^2 ))))  , with  m^2 =(√((9/4)((a^2 /b^2 )−1)^2 +(a^2 /b^2 ))) −(3/2)((a^2 /b^2 )−1) .
$${Minimum}\:{Area}\left(\bigtriangleup{ABC}\:\right)=\:\bigtriangleup_{{min}} \\ $$$$\bigtriangleup_{{min}} =\frac{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} }{{m}\left(\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)}\:\:,\:{with} \\ $$$${m}^{\mathrm{2}} =\sqrt{\frac{\mathrm{9}}{\mathrm{4}}\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}\:−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)\:. \\ $$
Commented by ajfour last updated on 11/Nov/18
See Q. 47559  for solution.
$${See}\:{Q}.\:\mathrm{47559}\:\:{for}\:{solution}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *