00:01
In this problem we are provided with an ellipse 1 over 4 times x squared plus y squared equals to 1 and we are given that we need to inscribe a triangle into the ellipse such that 1 vertex of the triangle is at negative 2 comma 0 and the opposite side is perpendicular to the x axis.
00:34
So we need to find out the largest possible area of the triangle.
00:48
So now in order to do so, we must first find out an expression for the area of the triangle.
00:55
So first let us sketch out the given information.
01:00
From the figure, we have the vertices of the triangle to b.
01:04
A is negative 2 .0.
01:06
B is 2 times cost theta, sine theta and c is 2 times sine theta, sorry, 2 times cost theta, negative sine theta.
01:25
So now in order to find out the area of the triangle, we require the length of the base, which is bc, and the length of the height.
01:39
Let us denote the height by h.
01:41
So here we have the height h to be equal to 2 times 1 plus cos theta and we have bc to be equal to 2 times sine teta.
01:54
So substituting this we have the value of a as 1 over 2 times 2 sine theta times 2 times 1 plus cos theta.
02:07
So simplifying this we see that 2 and 2 get cancelled and 2.
02:11
And we are left with 2 times sine theta times 1 plus cos theta.
02:17
So now in order to find out the maximum area, we differentiate a with respect to theta, we obtain da over d theta to be equal to making use of the product rule we have 2 times sine theta times negative sine theta plus 2 times cos theta times 1 plus cos theta.
02:44
So next let us simplify this.
02:50
So on simplification we get negative 2 times sine squared theta plus 2 times cos theta times 1 plus cos theta.
03:05
So we know that sine square theta can be written as 1 minus cos square theta plus 2 times cos square theta plus 2 times cos tase 2 times cos tata...