Answer:
[tex]\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1[/tex]
Step-by-step explanation:
Let P be the moving point and the given points be A (c,0) and B (-c, 0). If (h, k) be the co-ordinates of any position of P on its locus then:
PA + PB = 2a
PA = 2a - PB
PA² = 4a² + PB² – 4a ∙ PB
PA² – PB² = 4a² – 4a ∙ PB
[(h - c)² +(k - 0)²] - [(h + c)² +(k - 0)²] = 4a² – 4a. PB
-4hc = 4a² – 4a∙PB
a ∙ PB = a² + hc
Taking square
a² ∙ PB² = (a² + hc)²
a² [(h + c)² + (k - 0)²] = (a² + hc)²
a² [h² + c² + 2hc + k²] = a⁴ + 2a²hc + h²c²
a²h² – h²c² + a²k² = a⁴ – a²c²
(a² – c²)h² + a²k² = a² (a² – c²)
h²/a² + k²/(a² – c²) = 1
Therefore, the required equation to the locus of P is
[tex]\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1[/tex]
