A problem for discussion - Find the domain of a constant and the equation of a line?

Question

A and B are the two points on the curve 3x^2 + y^2 = c. N(1,3) is the mid point of segment AB. The perpendicular bisector of segment AB intersects the curve at points C and D.

1)Find the domain of c and the equation of the line AB.
2)What value of c can guarantee that points A, B, C, and D are on the same circle?

gjmb1960,

Try c = 1. It doesn't work.

Answer 1

1) c>12. Equation of AB is y+x = 4.
2) Any c in the aforementioned domain will make A,B,C,D on a circle with a center at the point (-1/2,3/2) and a radius of a length of sqrt(2c-6)/2.
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Proof of (1):
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Since (1,3) is the mid point of two points lying on the graph of an ellipse(3x^2+y^2=c is an ellipse equation), this point must lie inside the ellipse because the ellipse is a convex shape.

Thus, 3(1)^2+(3)^2=3+9=12
So, the domain is c>12.
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Since N(1,3) is the mid point of A(x1,y1)&B(x2,y2), we have:
(x1+x2)/2 = 1, (y1+y2)/2=3 or
x1=2-x2, y1=6-y2

Now, A&B lie on the ellipse and hence satisfy the equation of this ellipse. Substitute their coordinates and use the last equation:

3(2-x2)^2+(6-y2)^2=c or, after simplification,
3(x2)^2+(y2)^2+(48-12x2-12y2)=c

3(x2)^2+(y2)^2=c

Subtract the last equation from its preceding one to get:
48-12x2-12y2=0 or, upon dividing through by 12,

x2+y2=4

Now, use x1=2-x2, y1=6-y2 to find that
x1+y1=(2-x2)+(6-y2) =8-(x2+y2)=8-4=4

so x1+y1 = 4 = x2+y2

Thus, the line through A&B is x+y = 4
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Proof of (2):
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Up to now, we have found relations between A&B as follows:

x1=2-x2, y1=6-y2
x1+y1=4, x2+y2=4

Let's now find the forms of the other two points of intersection of the perpendicular bisector of the previous line with the ellipse.
This will help us know what values of c, if any, will lead to a circle lying on A,B,C,D points.

The equation of AB is y=4-x, so the equation of its perpendicular bisector is y = 3+a(x-1) since this perpendicular bisctor bisects the segment AB or passes through its middle, which is N(1,3).

Now, we need the slope of the perpendicular.
It is simply -1/slope of AB or -1/-1=1=a.
So we have its equation as:
y=3+(x-1) or

y=2+x
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Now, let's find the points of intersection of the perpendicular bisector with the ellipse.
Substitute y=2+x into the equation of the ellipse to get:

3x^2+(2+x)^2=c or
4x^2+4x+4-c =0. Now, solve using the quadratic formula:
x=[-4+-sqrt(16-16(4-c)]/[2*4] or, upon simplification,we find the points C&D as:

x=[-1+-sqrt(c-3)]/2 & using y=2+x
y=[3+-sqrt(c-3)]/2

similarly, you can find that the A,B points in terms of c are:
x=[2+-sqrt(c-12)]/2 & using y=4-x
y=[6-+sqrt(c-12)]/2
(Notice that signs of the sqrt are reversed for y, which means the +sqrt for x goes with the -sqrt for y, and vice verse.

Since the CD line is the perpendicular bisector of AB, any circle passing through all these points will have its center on CD.
Now, it remains to determine where on the line CD this center lies.
To this end, remember that C&D are points on the circle and on one line, and conclude that the center of the circle is the mid point of the line segment CD. Thus,

([-1+sqrt(c-3)]/2 + [-1-sqrt(c-3)]/2)/2 = -1/2 &
([3+sqrt(c-3)]/2 + [3+sqrt(c-3)]/2)/2 = 3/2

So the center is always (-1/2,3/2).
This center does not depend on c, where c >12.

So for all c in the domain mentioned in part 1, there is a circle passing through all points A,B,C,D and having its center at
(-1/2,3/2).
The radius, however, does depend on c. It can be calculated as the distance between the center (-1/2,3/2) and one of the points A,B,C,D.

r = sqrt[([-1+sqrt(c-3)]/2-[-1/2])^2 + ([3+sqrt(c-3)]/2-[3/2])^2]
= sqrt(2c-6)/2, upon simplification.

Answer 2

Hey. I just wanted to tell you that I liked your answer ti the integral question.

I am reading your question. If c is positive, then you have an ellipse. Is this what you mean when you ask for the domain?

The semiaxes are sqrt c/3 and sqrt c. I suggest you to do the drawing to think this problem.

Ana

Answer 3

1) c>0 ( otherwise ther would be no midpoint )
2) 4