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प्रश्न
The coordinates of a moving point P are `("a"/2 ("cosec" theta + sin theta), "b"/2 ("cosec" theta - sin theta))` where θ is a variabe parameter. Show hat the equation of the locus P is b2x2 – a2y2 = a2b2
उत्तर
Let P(h, k) be the moving point
We are given P = `["a"/2 ("cosec" theta + sin theta), "b"/2 ("cosec"theta - sin theta)]`
⇒ h = `"a"/2 ("cosec theta + sin theta)`
(i.e.) a[cosec θ + sin θ] = 2h
⇒ cosec θ + sin θ = `(2"h")/"a"` ......(1)
`"b"/2 ("cosec"theta - sintheta)` = k
⇒ cosec θ – sin θ = `(2"k")/"b"` ......(2)
(1) + (2) ⇒ 2cosec θ = `(2"h")/"a" + (2"k")/"b"`
(÷ by 2) ⇒ cosec θ = `"h"/"a" + "k"/"b"` ......(3)
(1) – (2) ⇒ 2 sin θ = `(2"h")/"a" - (2"k")/"b"`
(÷ by 2) ⇒ sin θ = `"h"/"a" - "k"/"b"` ......(4)
But cosec θ sin θ = 1
So from (3) and (4)
⇒ `("h"/"a" + "k"/"b") ("h"/"a" - "k"/"b")` = 1
⇒ `"h"^2/"a"^2 - "k"^2/"b"^2` = 1
So the locus of (h, k) is `x^2/"a"^2 - y^2/"b"^2` = 1
(i.e) `("b"^2x^2 - "a"^2y^2)/("a"^2"b"^2)` = 1
⇒ b2x2 – a2y2 = a2b2
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