Equation that represents all points in the plane & the sum distances from the fixed points (-3,2) (7,2) is 26?

Question

possible equations....



a. (x - 4)^2 /25 + (y - 3)^2 /16 = 1

b. (x - 1)^2 /16 + y^2 /25 = 1

c. (x - 2)^2 /169 + (y - 2)^2 /144 = 1

d. (x - 1)^2 /144 + (y + 1)^2 /169 = 1

e. (x - 1)^2 /25 + (y - 4)^2 /16 = 1

f. (x + 2)^2 /16 + (y - 1)^2 /25 = 1

g. (x - 2)^2 /25 + (y - 2)^2 /16 = 1

or none of these

Answer 1

The set of all points whose distances from two fixed points is a constant, is an ellipse. So (-3,2) and (7,2) are the two focus points. The center is always right in the middle of the two focus points so the ellipse's center is ((-3+7)/2, 2) = (2,2). This eliminates the choices to c or g.

Consider right end point of the elipse. It's on the y=2 line, so it's in the form of (x,2). The distance between this and (-3,2) is x - (-3) = x+3. The distance between the end point and (7,2) is x - 7. Adding these up gives 2x - 4 = 26, so x = 15. This means (15,2) should be a point on the ellipse. You'll find that this point works for one our two possible answers but not the other one.

Answer 2

Find the equation of the locus of points in the plane, the sum of whose distances from the fixed points (-3,2) and (7,2) is 26.

This is the definition of an ellipse. The two given points are the foci. The center (h, k) of the ellipse is the midpoint between the two foci.

(h, k) = (2, 2)

The foci are in a horizontal line so the major axis of the ellipse is horizontal.

The distance between the two foci is 2c.
2c = 7 - -3 = 10
c = 5
c² = 25

The sum of the distances from a point on the ellipse to the two foci is 2a.
2a = 26
a = 13
a² = 169

b² = a² - c² = 169 - 25 = 144

The equation of the ellipse is:

(x - h)²/a² + (y - k)²/b² = 1

Plug in the values.

(x - 2)²/169 + (y - 2)²/144 = 1

The answer is c.