8) Find the maximum area of an isosceles triangle inscribed in the ellipse \frac{x ^2 }{a^2}+ \frac{y^2}{b^2 } = 1 with its vertex at one end of the major axis.

Answers (1)


Given the equation of the ellipse
\frac{x ^2 }{a^2}+ \frac{y^2}{b^2 } = 1
Now, we know that ellipse is symmetrical about x and y-axis. Therefore, let's assume coordinates of A = (-n,m) then,
Now,
Put(-n,m) in equation of ellipse
we will get
m = \pm \frac{b}{a}.\sqrt{a^2-n^2}
Therefore, Now
Coordinates of A = \left ( -n,\frac{b}{a}.\sqrt{a^2-n^2} \right )
Coordinates of B = \left ( -n,-\frac{b}{a}.\sqrt{a^2-n^2} \right )
Now,
Lenghth AB(base) = 2\frac{b}{a}.\sqrt{a^2-n^2}
And height of triangle ABC = (a+n)
Now,
Area of triangle =  \frac{1}{2}bh
A = \frac{1}{2}.\frac{2b}{a}.\sqrt{a^2-n^2}.(a+n)= ab\sqrt{a^2-n^2}+bn\sqrt{a^2-n^2}
Now,
\frac{dA}{dn} = \frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{bn^2}{\sqrt{a^2-n^2}}
Now,
\frac{dA}{dn} =0\\ \frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{bn^2}{\sqrt{a^2-n^2}} =0\\ -abn + n(a^2-n^2)-bn^2 = 0\\ \Rightarrow n = -a,\frac{a}{2}
but n cannot be zero 
therefore, n = \frac{a}{2}
Now, at  n = \frac{a}{2}
\frac{d^2A}{dn^2}< 0
Therefore, n = \frac{a}{2} is the point of maxima
Now,
 b = 2\frac{b}{a}.\sqrt{a^2- (\frac{a}{2})^2}= \sqrt3b
h = (a+ n ) = a+ \frac{ a}{2} = \frac{ 3a}{2}
Now,
Therefore,  Area (A) = \frac{1}{2}bh = \frac{1}{2}\sqrt3 b \frac{3a}{2} = \frac{3\sqrt3ab}{4}
 

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