Question

find the locus of a point which moves so that the sum of its distance from (3,0) and (-3,0) is 9​

Answer 1

Answer:

(+2)2+(−3)2=9[(−0)2+(−3)2].⇒(x+2)2+(y−3)2=9[(x−0)2+(y−3)2].

⇒2+4+4+2−6+9=92+92−54+81⇒x2+4x+4+y2−6y+9=9x2+9y2−54y+81

⇒82−4+82−48+68=0.⇒8x2−4x+8y2−48y+68=0.

⇒82−4+12+82−48+72−92=0.⇒8x2−4x+12+8y2−48y+72−92=0.

⇒8(2−12+116)+8(2−6+9)=92.⇒8(x2−12x+116)+8(y2−6y+9)=92.

⇒(2−12+116)+(2−6+9)=916.⇒(x2−12x+116)+(y2−6y+9)=916.

⇒(−14)2+(−3)2=(34)2.⇒(x−14)2+(y−3)2=(34)2.

⇒⇒ The equation of the locus is  (−14)2+(−3)2=(34)2.(x−14)2+(y−3)2=(34)2.

Step-by-step explanation: