Advertisements
Advertisements
प्रश्न
Show that the set of all points such that the difference of their distances from (4, 0) and (– 4, 0) is always equal to 2 represent a hyperbola.
Advertisements
उत्तर
Let P(x, y) be any point.
We have `sqrt((x - 4)^2 + (y - 0)^2) - sqrt((x + 4)^2 + (y - 0)^2)` = 2
⇒ `sqrt(x^2 + 16 - 8x + y^2) - sqrt(x^2 + 16 + 8x + y^2)` = 2
Putting the x2 + y2 + 16 = z ......(i)
⇒ `sqrt(z - 8x) - sqrt(z + 8x)` = 2
Squaring both sides, we get
⇒ `z - 8x + z + 8x - 2sqrt((z - 8x)(z + 8x))` = 4
⇒ `2z - 2sqrt(z^2 - 64x^2)` = 4
⇒ `z - sqrt(z^2 - 64x^2)` = 2
⇒ `(z - 2) = sqrt(z^2 - 64x^2)` = 2
⇒ `(z - 2) = sqrt(z^2 - 64x^2)`
Again squaring both sides, we have
z2 + 4 – 4z = z2 – 64x2
⇒ 4 – 4z + 64x2 = 0
Putting the value of z, we have
⇒ 4 – 4(x2 + y2 + 16) + 64x2 = 0
⇒ 4 – 4x2 – 4y2 – 64 + 64x2 = 0
⇒ 60x2 – 4y2 – 60 = 0
⇒ 60x2 – 4y2 = 60
⇒ `(60x^2)/60 - (4y^2)/60` = 1
⇒ `x^2/1 - y^2/15` = 1
Which represent a hyperbola.
Hence proved.
