   Chapter 14, Problem 7P

Chapter
Section
Textbook Problem

If the ellipse x2/a2 + y2/b2 = 1 is to enclose the circle x2 + y2 = 2y, what values of a and b minimize the area of the ellipse?

To determine

To find: The value of a and b that minimizes the area of ellipse enclosing a circle.

Explanation

Given:

The equation of ellipse is x2a2+y2b2=1 (1)

The equation of circle is x2+y2=2y (2)

Calculation:

Rearrange the equation of circle,

x2+y22y=0

Add 1 on both sides.

x2+y22y+1=0+1

Apply the formula a22ab+b2=(ab)2 in above equation.

x2+(y1)2=1

Here, the circle has the center of (0,1) and the radius 1 . If the circle lies on x axis, the ellipse will intersect the circle at the any value of y . That y value must satisfy both the equations of ellipse and circle.

From equation (2)

x2=2yy2

Substitute 2yy2 for x2 in equation (1)

2yy2a2+y2b2=12ya2y2a2+y2b2=12ya2+y2b2y2a2=12ya2+y2(1b21a2)=1

2ya2+y2(a2b2b2a2)=1

The above equation in the form of Quadratic equation ay2+by+c=0 ,

y2(a2b2b2a2)+2a2y1=0 (3)

The solution for quadratic equation ay2+by+c=0 is y=b±b24ac2a , where the discriminant b24ac must be zero to have exactly one root as a real number.

Compare the equation (3) with quadratic equation,

a=a2b2b2a2 , b=2a2 and c=1 .

Substitute a2b2b2a2 for a , 2a2 for b and 1 for c in above equation.

b24ac=0(2a2)24(a2b2b2a2)(1)=04a4+4(a2b2b2a2)=04a2[1a2+a2b2b2]=01a2+a2b2b2=0

b2+a2(a2b2)a2b2=0b2+a2(a2b2)=0b2+a4a2b2=0b2a2b2+a4=0 (4)

Consider, b2a2b2+a4=g(a,b)

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