Triangle Inequality and Locus in Coordinate Geometry
A geometry problem asks you to find the locus of points P in the plane such that the sum of distances from P to two fixed points A = (-3, 0) and B = (3, 0) is 10. A student immediately says the locus is a circle of radius 5 centered at the origin, because the midpoint of AB is the origin and the sum of distances is constant. You recognize this as an ellipse, but you need to justify the parameters and check edge cases. You also need to determine whether the locus includes all points satisfying the condition, and whether the student's circle is entirely contained in the locus, intersects it, or is disjoint. You decide to derive the equation from the definition and compare with the circle x^2 + y^2 = 25.
1. Derive the Cartesian equation of the locus and identify its type, center, and axes. Show that it is not a circle.
Read the worked answer
Let P = (x, y). The condition is sqrt((x+3)^2 + y^2) + sqrt((x-3)^2 + y^2) = 10. This is the definition of an ellipse with foci A and B, major axis length 2a = 10 so a = 5, and focal distance c = 3 (since the distance between foci is 6, c = 3). Then b^2 = a^2 - c^2 = 25 - 9 = 16, so b = 4. The center is the midpoint of AB, which is (0,0). The major axis is along the x-axis because the foci lie on the x-axis. The equation is x^2/25 + y^2/16 = 1. This is not a circle because the denominators 25 and 16 are different; the semi-axes are 5 and 4. The student's circle x^2 + y^2 = 25 has radius 5, which matches the semi-major axis but not the semi-minor axis. The ellipse is inscribed in the circle? Actually the ellipse lies inside the circle x^2 + y^2 = 25 because for points on the ellipse, x^2/25 + y^2/16 = 1 implies x^2 + y^2 = 25 - (9/16)y^2 ≤ 25, with equality only when y=0. So the ellipse is contained in the circle and touches it at (±5,0).
2. Does the circle x^2 + y^2 = 25 contain any points that satisfy the distance-sum condition? Justify by testing a point on the circle not on the x-axis.
Read the worked answer
Take a point on the circle, say (0,5). Its distance to A = (-3,0) is sqrt(9+25)=sqrt(34)≈5.831, and to B=(3,0) is also sqrt(34). The sum is 2√34 ≈ 11.662, which is greater than 10. So (0,5) does not satisfy the condition. In fact, for any point on the circle x^2+y^2=25 with y≠0, the sum of distances to A and B is strictly greater than 10. Proof: The ellipse x^2/25 + y^2/16 = 1 is the set of points with sum exactly 10. For a point on the circle, x^2+y^2=25, substitute into the ellipse equation: x^2/25 + y^2/16 = (25 - y^2)/25 + y^2/16 = 1 - y^2/25 + y^2/16 = 1 + y^2(1/16 - 1/25) = 1 + y^2(9/400). This is >1 when y≠0. Since the ellipse equation equals 1 exactly for sum=10, and the function is increasing in the sum, a value >1 means the sum is >10. Thus the circle and ellipse intersect only at (±5,0). The student's locus is incorrect; the correct locus is the ellipse.