Question -
Answer -
Let P (h, k) be any point on the locus and let A (ae, 0) and B(-ae, 0).
Where, PA – PB = 2a
Now again let us square on both the sides we get,
(eh + a)2 = (h +ae)2 + (k – 0)2
e2h2 + a2 +2aeh = h2 + a2e2 +2aeh + k2
h2 (e2 – 1) – k2 =a2 (e2 –1)
Now let us replace (h, k) with (x, y)
The locus of a point such that the difference of its distancesfrom (ae, 0) and (-ae, 0) is 2a.
Where b2 = a2 (e2 –1)
Hence proved.