Circumscribing an Ellipse Let P(x, a) and points on the upper half of the ellips

Question

Circumscribing an Ellipse Let P(x, a) and points on the upper half of the ellipse x^2/100 + (y - 5)^2/25 = 1 centered at (0, 5). A triangle RST is formed by using lines to the ellipse at Q and P as shown in the figure Show that the area of the triangle is A(x) = -f'(x)[x - f(x)/f'(x)]^2, where y = f(x) is the function representing the upper half of the ellipse. What is the domain of A? Draw the graph of A. How are asymptotes of the graph related to the problem situation? Determine the height of the triangle with minimum area How is it related to the y-coordinate of the center of the Repeat parts (a)-(c) for the ellipse x^2/C^2 + (y - B)^2/B^2 = 1 centered at (0, B). Show that the triangle has minimum area when its height is 3B.

Explanation / Answer

We can see from the given figure that the base of the triangle RST is represented by x-axis.

Let the coordinates of point R be (0, b).

It is given that y = f(x) is the function representing the upper half of the ellipse, both the tangents RS & RT are symmetrical about y-axis, so the slopes of tangents will be equal in magnitude and opposite in sign.

The slope of tangent = dy/dx = f'(x)

Now by point-slope equation the slope of tangent passing through two points = (y2 - y1) / (x2 - x1)

So, f'(x) = b/x2 & -b/x2 [ here x2 is x-coordinate of S & T ]

Now we get in this way : S (-b/f'(x), 0) & T (b/f'(x), 0)

Therefore, area of RST = 1/2 * B * H = 1/2 * 2b/f'(x) * b

or area = b2/f'(x) ........ (1)

Now, in order to eliminate 'b' from area (1), we will find the slope of the tangent line from R (0, b) to P (x, a) is

f' = (a - b)/x, which gives us, b = a - x*f'(x)

Now since 'a' is y-coordinate of both the given points lying on y = f (x) curve of ellipse, so we can write f (x) = a, thus

b = f(x) - x*f'(x)

putting this value of 'b' in (1), we get

Area = [{ f(x) - x*f'(x) }2 ]/f'(x)

Area = [{ f'(x){f(x)/f'(x) - x}}2 ]/f'(x)

Area = f'(x){f(x)/f'(x) - x}2

Area = - f'(x){ x - f(x)/f'(x) }2 .......... Proved

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