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  • The area of the greatest isosceles triangle that can be inscribed in the ellipse <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

The area of the greatest isosceles triangle that can be inscribed in the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} having its vertex coincident with one extremity of major axis, is

Option: 1

\mathrm{\frac{3}{4} a b}


Option: 2

\mathrm{\frac{1}{2}(3 \sqrt{3}) a b}


Option: 3

\mathrm{\frac{1}{4}(3 \sqrt{3}) a b}


Option: 4

\mathrm{\frac{1}{4}(9 \sqrt{3}) a b}


Answers (1)

best_answer

Let APQ be the isosceles triangle inscribed in the ellipse \mathrm{x^2 / a^2+y^2 / b^2=1} with one vertex A at the extremity (a, 0) of the major axis. If coordinates of P are \mathrm{(a \cos \theta, b \sin \theta)}, then those of Q will be \mathrm{(a \cos \theta,-b \sin \theta).}

So, area of \mathrm{\triangle A P Q} is given by

\mathrm{ \begin{aligned} & \Delta=\frac{1}{2} P Q \cdot A D=P D \cdot A D=P D \cdot(O A-O D) \\ & =b \sin \theta(a-a \cos \theta)=\frac{1}{2} a b(2 \sin \theta-\sin 2 \theta) \end{aligned} }
Now, \mathrm{\frac{d \Delta}{d \theta}=a b(\cos \theta-\cos 2 \theta)=0}

\mathrm{ \Rightarrow \cos \theta=\cos 2 \theta \Rightarrow \theta=2 \pi / 3 }

Now, \mathrm{ \frac{d^2 \Delta}{d \theta^2}=a b(-\sin \theta+2 \sin 2 \theta)}

When \mathrm{\theta=2 \pi / 3, \frac{d^2 \Delta}{d \theta^2}=-\frac{1}{2}(3 \sqrt{3}) a b<0}

Thus, \mathrm{\Delta} has maximum value when \mathrm{\theta=2 \pi / 3.}
\mathrm{ \text { Therefore, maximum } \begin{aligned} \Delta & =\frac{1}{2} a b\left[2 \sin \left(\frac{2 \pi}{3}\right)-\sin \left(\frac{4 \pi}{3}\right)\right] \\ & =\frac{1}{4}(3 \sqrt{3}) a b \end{aligned} }

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