8) Find the maximum area of an isosceles triangle inscribed in the ellipse with its vertex at one end of the major axis.

Name: Find the maximum area of an isosceles triangle inscribed in the ellipse x raised to 2 by a raised to 2 plus y raised 2 by b raised to 2 is equal to 1 with its vertex at one end of the major axis.
Uploaded: Mon, 2019-06-10 13:15
Description: Find the maximum area of an isosceles triangle inscribed in the ellipse x raised to 2 by a raised to 2 plus y raised 2 by b raised to 2 is equal to 1 with its vertex at one end of the major axis.

Find the maximum area of an isosceles triangle inscribed in the ellipse x raised to 2 by a raised to 2 plus y raised 2 by b raised to 2 is equal to 1 with its vertex at one end of the major axis.

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) $= \frac{1}{2}bh = \frac{1}{2}\sqrt3 b \frac{3a}{2} = \frac{3\sqrt3ab}{4}$