 Hello and welcome to the session. Let's work out the following problem. It says find the maximum area of an isosceles triangle inscribed in ellipse x square upon a square plus y square upon v square is equal to 1 with its vertex at one end of the major axis. So let's first see the figure. We have to find the maximum area of the isosceles triangle inscribed in ellipse with its one end, with its one vertex at the one end of major axis. So this is a zero, this point is a zero and any point on the ellipse in the first quadrant will be given by a cos theta b sin theta, right? So this much distance is b sin theta and this much distance is b cos theta, right? So let's now move on to the solution and let a p p dash be the isosceles triangle inscribed in ellipse with one end, the point a and the coordinates are a zero since it lies on major axis, right? Now the area of triangle, the isosceles triangle, a p p dash will be 1 by 2 into base into height. Now base is p p dash and height is a m. So the area of triangle a p p dash will be given by 1 by 2 into p p dash into a m. Now p p dash is twice of p m and p m is b sin theta as this length is b sin theta. So this is equal to 2 b sin theta and a m is equal to a o minus o m, this distance minus this. Now a o is a since this much distance is a and o m is a cos theta as we discussed earlier that this distance is a cos theta that is minus a cos theta. Therefore the area of the triangle let's denote it by capital A. So area is given by 1 by 2 into base that is 2 b sin theta into a into 1 minus cos theta. This is further equal to 1 by 2 into 2 a b sin theta minus 2 a b cos theta sin theta or sin theta cos theta. Now taking a b common we have 1 by 2 into a b into 2 sin theta minus 2 sin theta cos theta which can be written as sin 2 theta as we know that 2 into sin theta cos theta is sin 2 theta. Now this is a now we have to find d a by d theta for the maximum area. So d a by d theta would be equal to 1 by 2 into a b into derivative of 2 sin theta is 2 cos theta derivative of sin 2 theta is 2 cos 2 theta. Now for maximum or minima we put d a by d theta equal to 0. So we have d a by d theta which is equal to 1 by 2 into a b into 2 cos theta minus 1 by 2 into a b into 2 cos 2 theta equal to 0. So we have a b into cos theta minus cos 2 theta equal to 0. So this implies cos theta minus cos 2 theta equal to 0 because a b the product cannot be 0. So this implies cos theta is equal to cos 2 theta. Now again cos theta can be written as cos 2 pi minus theta is equal to cos 2 theta. So this implies 2 pi minus theta is equal to 2 theta. So this implies 3 theta is equal to 2 pi and this implies theta is equal to 2 pi by 3. Now we will see that the second derivative of the area with respect to theta is less than 0. So d square a by d theta square will be equal to d a by d theta is this cos theta a b into cos theta minus cos 2 theta. So it would be a b into minus sin theta into minus 2 sin 2 theta since the derivative of cos 2 theta is minus 2 sin 2 theta. So here we will be having positive sign. Now we find the value of d square a by d theta square when theta is 2 pi by 3. So this is equal to a b into minus sin 2 pi by 3 plus 2 sin 2 theta. So it would become 4 pi by 3. Now this is again equal to a b into sin 2 pi by 3 is root 3 by 2. So this is minus root 3 by 2 and sin 4 pi by 3 is minus 3 pi by 2. So this becomes minus 2 into root 3 by 2 and this whole quantity comes out to be less than 0. Hence area is maximum when theta is 2 pi by 3 and the maximum area is given by now a is this 1 by 2 into a b into 2 sin theta minus sin 2 theta. So now we will put theta equal to pi by 2 pi by 3. So it becomes 1 by 2 into a b into sin 2 pi by 3 minus 4 pi by 3 and here we have 2 sin 2 theta and this is equal to 1 by 2 into a b into 2 sin 2 pi by 3 is root 3 by 2 sin 4 pi by 3 is minus root 3 by 2. So it becomes 3 by 2. So this is further equal to 1 by 2 into a b into 3 root 3 by 2 2 root 3 by 2 plus root 3 by 2 is 3 root 3 by 2. So this is equal to a b into 3 root 3 by 4. Hence the maximum area is 3 root 3 by 4 into a b. So this completes the question and the session. Bye for now. Take care. Have a good day.