Question

Find the equation of the set of all points, the sum of whose distances from the points (3,0) and (9,0) is 12

Answer 1

3x2+4y2−36x=03x2+4y2−36x=0
Let S(3,0)S(3,0) and S1(9,0)S1(9,0) be the two foci and P(x,y)P(x,y) be the moving point.
It is given that the sum of the distances from the point (3,0)(3,0) and (9,0)(9,0) is 1212.
∴(x−9)2+y2−−−−−−−−−−−√+(x−3)2+y2−−−−−−−−−−−√=12∴(x−9)2+y2+(x−3)2+y2=12
(ie) (x−9)2+y2−−−−−−−−−−−√=12−(x−3)2+y2−−−−−−−−−−−√(x−9)2+y2=12−(x−3)2+y2
Squaring on both sides we get,
(x−9)2+y2=144+(x−3)2+y2−24(x−3)2+y2−−−−−−−−−−−√(x−9)2+y2=144+(x−3)2+y2−24(x−3)2+y2
On simplifying we get,
12x+72=24(x−3)2+y2−−−−−−−−−−−√12x+72=24(x−3)2+y2
(ie) x+6=2(x−3)2+y2−−−−−−−−−−−√x+6=2(x−3)2+y2
Squaring once again we get,
(x+6)2=4[(x−3)2+y2](x+6)2=4[(x−3)2+y2]
On simplifying we get,
3x2+4y2−36x=0

Answer 2

Answer:3x^2+4y^2−36x=0

Step-by-step explanation:

t is given that the sum of the distances from the point (3,0) and (9,0) is 12.

Let the point be (x, y)

Distance between the point of intersection & centre =  [Distance Formula]

Squaring both the sides,

Squaring both the sides,

x2 + 12x + 36 = 4x2 - 24x + 36 + 4y + 2

3x2 - 36x + 4y2=0

Hence, the required equation is 3x2 - 36x + 4y2 = 0.